This is an arithmetic progression based question from Chapter name- Arithmetic progressions

In this question we have been given that there be an A.P.

first term a and common difference d.

Also, If an denotes its nth term and Sn is the sum of first n terms,

Now we have to find k, if Sn = 3n2 + 5n and ak = 164

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Understanding CBSE Mathematics

Class :- 10th

Question no 56(vii)

Given S

_{n}= 3n^{2}+ 5n,On putting n = 1, we get the first term(a), S

_{1}= a = 3(1)^{2 }+ 5(1) = 8On putting n = 2 gives S

_{2}= a + a + d = 3(2)^{2 }+ 5(2) = 22=> d = 22 – 2a

=> d = 22–16 = 6

The kth term of the A.P., a

_{k}= a + (k – 1)d = 164=> 8 + (k – 1)6 = 164

=> 6k = 162

=> k = 27

Hence, the value of k is 27.