In an AP is the sum of its first n terms is 3n^2+5n and its kth term is 164 find the value of k.

In an AP is the sum of its first n terms is 3n^2+5n and its kth term i

1.	In an AP is the sum of its first n terms is 3n^2+5n and its kth term is 164 find the value of k.
Answer» S = 3n2\xa0+ 5n\xa0S1\xa0= a1\xa0= 3 + 5 = 8\xa0S2\xa0= a1\xa0+ a2= 12 + 10 = 22 ⇒ a2\xa0=\xa0S2\xa0-\xa0S1\xa0= 22 - 8 = 14\xa0S3\xa0= a1\xa0+ a2\xa0+ a3 = 27 + 15 = 42⇒a3\xa0=\xa0S3\xa0-\xa0S2\xa0= 42 - 22 = 20\xa0∴ Given AP is 8,14,20,.....\xa0Thus a = 8 , d = 6 Given tk= 164.\xa0164 = [a + (k -1)d]\xa0164 = [(8) + (k-1)6]\xa0164 = [8 + 6k - 6]\xa0164 = [2 + 6k]162= 6k, k= 162 / 6.\xa0∴ k = 27