This is an arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.6
In this question we have been given that the sum of first 9 terms of an A.P. is 162.
Also The ratio of its 6th term to its 13th term is 1: 2.
Now we have to Find the first and 15th term of the A.P
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Understanding CBSE Mathematics
Class :- 10th
Question no 28
We know sum of n terms of an A.P. is given by Sn = n[2a + (n − 1)d] / 2.
Therefore, Sum of first 9 terms of given A.P. = S9 = 9[2a + (9 − 1)d] / 2 = 162
=> 162 = 9(2a + (9 − 1)d) / 2
=> 2a + 8d = 36
=> a + 4d = 18 …..(1)
By using the formula of nth term of an A.P.
an = a + (n – 1)d
So,
Given a6 : a13 = 1 : 2,
=> a13 = 2a6
=> a+12d = 2(a + 5d)
=> a+12d = 2a + 10d
=> a = 2d …..(2)
On putting (2) in (1), we get,
=> 2d + 4d = 18
=> 6d = 18
=> d = 3
On putting d = 3 in (2), we get,
a = 2(3) = 6, which is the first term.
Now 15th term, a15 = a + 14d = 6 + 14 × 3 = 6 + 42 = 48
Hence, the first and 15th term of the A.P. are 6 and 48 respectively.