Summation MCQ Quiz - Objective Question with Answer for Summation - Download Free PDF
Last updated on Jun 15, 2025
Latest Summation MCQ Objective Questions
Summation Question 1:
If the first term is 27 and the common ratio is 2/3, what will be the 4th term of the GP?
Answer (Detailed Solution Below)
Summation Question 1 Detailed Solution
Given:
First term (a) = 27
Common ratio (r) = 2/3
Find the 4th term of the GP.
Formula used:
n-th term of GP = a × r(n-1)
Calculation:
4th term = 27 × (2/3)(4-1)
⇒ 4th term = 27 × (2/3)3
⇒ 4th term = 27 × (8/27)
⇒ 4th term = 8
∴ The correct answer is option (1).
Summation Question 2:
Let \(x=\frac{1}{5+\frac{1}{6+\frac{1}{5+\frac{1}{6+\ldots \ldots . . \infty}}}} .\) Which of the following equals x?
Answer (Detailed Solution Below)
Summation Question 2 Detailed Solution
\(x=\frac{1}{5+\frac{1}{6+x}}\)
\(\Rightarrow x=\frac{6+x}{31+5 x}\)
\(\Rightarrow 5 x^{2}+31 x=x+6\)
\(\Rightarrow 5 x^{2}+30 x-6=0\)
\(\Rightarrow x=\frac{-30 \pm \sqrt{900+120}}{10}=-3 \pm \sqrt{10.2}\)
The continued fraction has to be positive.
Therefore, We reject the negative value.
Summation Question 3:
The given series \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{5n - 1}}\)
Answer (Detailed Solution Below)
Summation Question 3 Detailed Solution
Concept:
A series in which the terms are alternatively positive or negative is called an alternating series.
Liebnitz’s series:
An alternating series u1 – u2 + u3 – u4 + … converges if
(i) Each term is numerically less than its preceding term,
(ii) \(\mathop {\lim }\limits_{n \to \infty } {u_n} = 0\)
If \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\), the given series is oscillatory.
Calculation:
Given series is \(\mathop \sum \limits_{n = 1}^\infty \frac{{{{\left( { - 1} \right)}^{n - 1}}n}}{{5n - 1}}\)
Now \({u_n} = \frac{n}{{5n - 1}} \)
⇒ \({u_n} - {u_{n - 1}} = \frac{n}{{5n - 1}} - \frac{{n - 1}}{{5n - 8}} = \frac{{ - 2n-1}}{{\left( {5n - 1} \right)\left( {5n - 8} \right)}}\) < 0
So each term is numerically less than its preceding term.
Now limit,
\(\mathop {\lim }\limits_{n \to \infty } {u_n} = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{5n - 1}} = \frac{1}{2}\)
⇒ \(\mathop {\lim }\limits_{n \to \infty } {u_n} \ne 0\)
∴ Series is oscillatory
Summation Question 4:
Test for convergence \(\rm \Sigma \frac{\sqrt{5n^2-5n+1}}{7n^3-7n^2+2} \)
Answer (Detailed Solution Below)
Summation Question 4 Detailed Solution
Given:
\(\rm \Sigma \frac{\sqrt{5n^2-5n+1}}{7n^3-7n^2+2} \)
Concept used:
Limit Comparision test:
if an and bn are two positive series such that \(\underset{n \rightarrow \infty}{L t} \frac{a_n}{b_n} = c \)
where c > 0 and finite then, either Both series converges or diverges together
P - Series test:
∑ \(\frac{1}{n^p }\)is convergent for p > 1 and divergent for p ≤ 1
Calculations:
nth term of the given series = un = \(\rm \Sigma \frac{\sqrt{5n^2-5n+1}}{7n^3-7n^2+2} \)
Let \(\rm v_n=\frac{1}{n^2}\)
\(\rm \displaystyle Lt_{n\rightarrow\infty}\frac{u_n}{v_n}=\displaystyle Lt_{n\rightarrow\infty}\left[\frac{n\sqrt{5-\frac{5}{n}+\frac{1}{n^2}}}{n^3\left(7-\frac{7}{n}+\frac{2}{n^3}\right)}\times\frac{n^2}{1}\right] \)
\(\rm =\displaystyle Lt_{n\rightarrow\infty}\left[\frac{\sqrt{5-\frac{5}{n}+\frac{1}{n^2}}}{\left(7-\frac{7}{n}+\frac{2}{n^3}\right)}\right]=\frac{\sqrt5}{7}\ne0 \)
∴ By comparison test, Σun and Σvn both converge or diverge.
But Σvn is convergent. [p series test - p = 2 > 1]
∴ Σun is convergent.
Summation Question 5:
Test for convergence \(\rm \displaystyle \Sigma_{n=1}^{\infty}\left(\frac{5^n+3}{6^n+1}\right)^{1/2}\)
Answer (Detailed Solution Below)
Summation Question 5 Detailed Solution
Given:
\(\rm \displaystyle \Sigma_{n=1}^{\infty}\left(\frac{5^n+3}{6^n+1}\right)^{1/2} \)
Concept used:
Limit Comparision test:
if an and bn are two positive series such that \(\underset{n \rightarrow \infty}{L t} \frac{a_n}{b_n} = c \) where c > 0 and finite then, either Both series converge or diverge together
P - Series test:
∑ \(\frac{1}{n^p }\)is convergent for p > 1 and divergent for p ≤ 1
\(\rm \frac{u_n}{v_n}=\left(\frac{1+\frac{3}{5^n}}{1+\frac{1}{6^n}}\right)^{1/2} \)
\(\rm \displaystyle Lt_{n\rightarrow \infty}\frac{u_n}{v_n}=1\ne0 \) ;
∴ By comparison test, Σun and Σvn behave the same way.
But Σvn = \(\rm Σ_{n=1}^{\infty}\left(\frac{5}{6}\right)^{n/2}=\sqrt{\frac{5}{6}}+\frac{5}{6}+\left(\frac{5}{6}\right)^{3/2}+.... \) which is a geometric series with common ratio \(\rm \sqrt{\frac{5}{6}}\) which is less than 1.
∴ Σvn is convergent.
Hence Σun is convergent.
Top Summation MCQ Objective Questions
If an AP is 13, 11, 9……, then find the 50th term of that AP.
Answer (Detailed Solution Below)
Summation Question 6 Detailed Solution
Download Solution PDFGiven,
The given AP is 13, 11, 9……
Formula:
Tnt = a + (n – 1)d
a = first term
d = common term
Calculation:
a = 13
d = 11 – 13
d = (-2)
T50 = 13 + (50 – 1) × (-2)
⇒ T50 = 13 + 49 × (-2)
⇒ T50 = 13 – 98
∴ T50 = -85
The sum of the series 1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ........... will be:
Answer (Detailed Solution Below)
Summation Question 7 Detailed Solution
Download Solution PDFConcept:
a + ar + ar2 + ar3 +…..
Sum of the above infinite geometric series:
\(=\frac{a}{1-r}\)
Analysis:
Given:
1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......
let x = (a2 + 1)
The series now becomes
S = 1 + 2x + 3x2 + 4x3 + ...... ----(1)
By multiplying x on both sides we get
xS = x + 2x2 + 3x3 + 4x4 + ...... ----(2)
Subtracting (1) and (2), we get
S(1 - x) = 1 + x + x2 + x3 + ..... ---(3)
The right hand side of (3) forms infinite geometric series with a = 1, r = x
∴ S(1 - x) = \(\frac{1}{1-x}\)
\(\Rightarrow S = \frac{1}{(1-x)^2}\)
putting the value of x, we get
\(\Rightarrow S = \frac{1}{(1- a^2 - 1)^2}\)
\(\Rightarrow S = \frac{1}{a^4}\)
What is the sum of the first 12 terms of an arithmetic progression if the first term is 5 and last term is 38?
Answer (Detailed Solution Below)
Summation Question 8 Detailed Solution
Download Solution PDFFormula used:
Sum of A.P. = n/2{first term + last term}
Calculation:
Number of terms = n = 12
⇒ Sn = 12/2{5 + 38}
⇒ Sn = 6{43}
⇒ Sn = 258The sum of first five multiples of 3 is
Answer (Detailed Solution Below)
Summation Question 9 Detailed Solution
Download Solution PDFGiven:
The first five multiples of 3
Concept:
Multiples = A multiple is a number that can be divided by another number a certain number of time without a remainder
Calculation:
⇒ The first five multiple of 3 = (3 × 1), (3 × 2), (3 × 3), (3 × 4), (3 × 5) = 3, 6, 9, 12, and 15
⇒ The sum of the multiple = 3 + 6 + 9 + 12 + 15 = 45
∴ The required result will be 45.
Identify the next number in the sequence.
1, 2, 4, 7, 11, _____
Answer (Detailed Solution Below)
Summation Question 10 Detailed Solution
Download Solution PDFThe pattern followed here is –
Hence 16 will complete the series.
Find the sum of given arithmetic progression 8 + 11 + 14 + 17 upto 15 terms
Answer (Detailed Solution Below)
Summation Question 11 Detailed Solution
Download Solution PDFShortcut Trick
Formula Used:
Average = (Sum of observations)/(Number of observations)
Last term = a + (n - 1)d
Calculation:
The above series is in arithmetic progression so the middlemost term 8th term will be the average
⇒ 8th term = 8 + (8 - 1) × 3 = 29
⇒ Sum of the series = 29 × 15 = 435
∴ The sum of the above series is 435
Additional Information
We can avoid this above (29 × 15) multiplication by digit sum Method and option
The digit sum of 29 is (2 + 9) ⇒ (11) ⇒ (2) and 15 is (1 + 5) = 6
⇒ 2 × 6 = 12 ⇒ (1 + 2) ⇒ 3
Now check the options whose digit sum will be 3 there is only option 2 whose Digit sum is 3
∴ 435 is the right answer
Traditional Method:
Given:
Arithmetic progression 8 + 11 + 14 + 17 upto 15 terms
Formula Used:
Sum of arithmetic progression = n[2a + (n - 1)d]/2
Calculation:
Sum of 1st 15 terms = 15[2 × 8 + (15-1)3]/2
⇒ (15 × 58)/2
⇒ 435
∴ 435 is the right answer
The sequence \(\left
Answer (Detailed Solution Below)
Summation Question 12 Detailed Solution
Download Solution PDFConcept:
The Nth term test
If \(\lim_{n\rightarrow ∞ }\left ( \sum_{n=0}^{∞ }a_{n} \right )=L\), where L is any tangible number other than zero. Then, \(\left ( \sum_{n=0}^{∞ }a_{n} \right )\) diverges.
This is also called the Divergence test.
Calculation:
We have, \(\left
\(\Rightarrow \sum_{n=1}^{∞ } log\left ( \frac{1}{n} \right )\)
\(\Rightarrow \lim_{n \to ∞ }\left [\sum_{n=1}^{∞ } log\left ( \frac{1}{n} \right ) \right ]\)
\(\Rightarrow \lim_{n \to ∞ }log\left ( \frac{1}{n} \right )\)
So, as n → ∞, \(\frac{1}{n}\) → 0
\(\Rightarrow \lim_{n \to ∞ }log\left ( \frac{1}{n} \right ) = -∞ \neq 0\)
Thus, our series diverges to -∞ by the nth term test.
Hence, The sequence \(\left
If S1, S2,.... Sn are the sums of n infinite geometrical series whose first terms are 1, 2, 3, .... n and common ratios are \(\frac{1}{2},\frac{1}{3}.\frac{1}{4},...,\frac{1}{{n + 1}}\) , then (S1 + S2 + S3 + ... + Sn) = ?
Answer (Detailed Solution Below)
Summation Question 13 Detailed Solution
Download Solution PDFGiven:
Series 1 = \( 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\)
Series 2 = \( 2+2\times\frac{1}{3}+2\times\frac{1}{3^2}+2\times\frac{1}{3^3}+...\)
...
Series n = \( n+n\times\frac{1}{n+1}+n\times\frac{1}{(n+1)^2}+n\times\frac{1}{(n+1)^3}+...\)
Concept:
Sum of an infinite G.P. series, S∞ = \( \frac{a}{1-r}\) , when |r| < 1
Sum of an A.P. series, SAP = \(\frac{n(a+l)}{2}\) , where, \(l\) = last term of series
Calculation:
For Series 1, a = 1, and r = \(\frac{1}{2}\)
∴ S1 = \(\frac{a}{1-r}\) = \(\frac{1}{1-\frac{1}{2}}\)
⇒ S1 = 2
Similarly, for Series 2, a = 2 and r = \(\frac{1}{3}\)
∴ S1 = \(\frac{2}{1-\frac{1}{3}}\)
⇒ S2 = 3
Similarly,
S3 = 4
S4 = 5, ...
Sn = n + 1
So, S1, S2, S3, ... , Sn is an Arithmetic Progression (A.P.),
For A.P.,
a = S1 = 2
n = n
\(l\) = Sn = n + 1,
Therefore,
(S1 + S2 + S3 + ... + Sn) = \( \frac{n(a+l)}{2} = \frac{n(2+n+1)}{2}\)
∴ ( S1 + S2 + S3 + ... + Sn ) = \(\frac{n(n+3)}{2}\)
What is the sum of the first 16 terms of an arithmetic progression if the first term is -9 and last term is 51?
Answer (Detailed Solution Below)
Summation Question 14 Detailed Solution
Download Solution PDFGIVEN:
First term a = – 9
Last term l = 51
Number of term n = 16
FORMULAE USED:
Sum = n/2 × (a + l)
CALCULATION:
⇒ 16/2 × (- 9 + 51) = sum
If ai > 0 for i = 1, 2, 3,..,n and a1, a2, a3, ...an = 1 then the greatest value of (1 + a1)(1 + a2)... (1 + an) is:
Answer (Detailed Solution Below)
Summation Question 15 Detailed Solution
Download Solution PDFLet the given expansion be f(n)
f(n) = (1 + a1)(1 + a2)... (1 + an)
Also given, a1 = a2 = a3 = ... = an = 1
Consider for n = 2
f(2) = (1 + a1)(1 + a2)
f(2) = (1 + 1)(1 + 1) = 22
Consider for n = 5
f(5) = (1 + a1)(1 + a2)(1 + a3)(1 + a4)(1 + a5)
f(5) = (1 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) = 25
Similary for n times, it is given as
f(n) = (1 + 1)(1 + 1) ...... (1 + 1) = 2n
(1 + a1)(1 + a2)... (1 + an) = 2n