Summation MCQ Quiz - Objective Question with Answer for Summation - Download Free PDF

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).

The continued fraction has to be positive.

Therefore, We reject the negative value.

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) 

If , the given series is oscillatory.

Calculation:

Given series is 

Now 

⇒  

So each term is numerically less than its preceding term.

Now limit,

⇒ 

∴ Series is oscillatory

Given:

Concept used:

Limit Comparision test:

if an and bn are two positive series such that  

where c > 0 and finite then, either Both series converges or diverges together

P - Series test: 

∑ is convergent for p > 1 and divergent for p ≤  1

Calculations:

n^th term of the given series = u_n = 

Let 

∴ By comparison test, Σu_n and Σv_n both converge or diverge.

But Σv_n is convergent. [p series test  - p = 2 > 1]

 ∴ Σu_n is convergent.

Given:

Concept used:

Limit Comparision test:

if an and bn are two positive series such that   where c > 0 and finite then, either Both series converge or diverge together

P - Series test: 

∑ is convergent for p > 1 and divergent for p ≤  1

 ;

∴ By comparison test, Σu_n and Σv_n behave the same way.

But Σv_n =  which is a geometric series with common ratio  which is less than 1.

∴ Σv_n is convergent.

Hence Σu_n is convergent.

Given,

The given AP is 13, 11, 9……

Formula:

T_nt = 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

Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

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) = 

putting the value of x, we get

Formula used:

Sum of A.P. = n/2{first term + last term}

Calculation:

Number of terms = n = 12

⇒ S_n = 12/2{5 + 38}

⇒ S_n = 6{43}

⇒ S_n = 258

Given:

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.

The pattern followed here is –

Hence 16 will complete the series.

Shortcut 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

⇒ 8^th 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

Concept:

The N^th term test

If , where L is any tangible number other than zero. Then,  diverges.

This is also called the Divergence test.

Calculation:

We have, \)

So, as n → ∞,  → 0

Thus, our series diverges to -∞ by the n^th term test.

Hence, The sequence \) is divergent to -∞.

Given:
Series 1 = 

Series 2 = 

...

Series n = 

Concept:
Sum of an infinite G.P. series, S_∞ =  , when |r| 

Sum of an A.P. series, S_AP =  , where,  = last term of series

Calculation:

For Series 1, a = 1, and r = 
∴ S₁ =  = 
⇒ S₁ = 2

Similarly, for Series 2, a = 2 and r = 
∴ S₁ = 
⇒ S₂ = 3

Similarly,

S3 = 4

S₄ = 5, ...

S_n = n + 1

So, S₁, S₂, S₃, ... , Sn is an Arithmetic Progression (A.P.),

For A.P.,

a = S1 = 2

n = n
 = Sn = n + 1,

Therefore,
(S1 + S2 + S3 + ... + S_n) = 

∴ ( S1 + S2 + S3 + ... + S_n ) = 

GIVEN:

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

∴ Sum = 336

Let 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