Q. 664.4( 15 Votes )

The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1:2. Find the first and 15th term of the A.P.

Answer :

Let a be the first term and d be the common difference

Now, a6 / a13 =


=


2a + 10d = a + 12d


a = 2d (i)


Now sum of first n terms of A.P


Sn = [2a + (n – 1) d]


S9 = [2a + 8d]


162 = 9 (a + 4d)


a + 4d = 18


2d + 4d = 18 [Using (i)]


6d = 18


d =3


Now from (i), we get


a = 2 * 3 = 6


So, first term, a1 = a = 6


15th term, a15 = a + 14d = 6 + 14(3)


= 6 + 42 = 48


Therefore, First term is 6 and 15th term is 48


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