Question

If the sum of n terms of an A.P. is $3n^{2} + 5n$ then which of its terms is 164?
(a) 26th
(b) 27th
(c) 28th
(d) none of these.

Answer 1

Answer:

27th term is 164.  

Among the given options option (b) 27th is correct.  

Step-by-step explanation:

Given :  

Sn = 3n² + 5n ………..(1)

On Putting n = 1 in eq 1,

S1 = 3(1)² + 5 × 1

S1 = 3 + 5

S1 = a1 = 8

On Putting n = 2 in eq 1,

S2 = 3(2)² + 5 × 2

S2 = 3 × 4 + 10

S2 = 12 + 10  

S2 = 22

On Putting n = 3 in eq 1,

S3 = 3(3)² + 5 × 3

S3 = 3 × 9 + 15

S3 = 27 + 15  

S3 = 42

⇒ a2 = S2 – S1  

⇒ a2 = 22 – 8  

⇒ a2 = 14

⇒ a3 = S3 – S2  

⇒ a3 = 42 – 22  

⇒ a3 = 20

∴ AP is 8, 14, 20, .....

Here,  a = 8, d = 14 - 8 = 6  

Given an = 164

By using the formula ,an = a + (n - 1)d

164 = 8 + (n –1) × 6

164 = (8) + (n –1)6

164 - 8 = 6n - 6

156 + 6 = 6n

162 = 6n

n = 162/6

n  = 27

Hence, 27th term is 164.  

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

(b) 27th term

Explanation:

Given

Sum of the n terms of the A.P. is

3n²+5n

sum of 1 term

S1 = 3(1)²+5(1)

S1 = 3+5

S1 = 8

sum of two terms

S2 = 3(2)²+5(2)

S2 = 3(4)+10

S2 = 12+10

S2 = 22

Sum of three terms

S3 = 3(3)²+5(3)

S3 = 3(9)+15

S3 = 27+15

S3 = 42

Now, let us find the terms of AP

First term

a = S1

a = 8

second term

a2 = S2 - S1

a2 = 22-8

a2 = 14

third term

a3 = S3-S2

a3 = 42-22

a3 = 20

We have series as

8, 14 , 20......

a = 14

d = a2 - a

d = 14-8

d = 6

an = 164

As we know

an = a+(n-1)d

On putting the values

164 = 8+(n-1)6

164 = 8+6n-6

164 = 2+6n

6n = 164-2

6n = 162

n = 162/6

n = 27

Hence, 164 is 27th term of the A.P.