Q16 of 176 Page 8

Find the 10th term of the A.P. whose 7th and 12th terms are 34 and 64 respectively.

We have


a7 = a + (7 – 1)d = a + 6d = 34 …(1)


and a12 = a + (12 – 1)d = a + 11d = 64 …(2)


Solving the pair of linear equations (1) and (2), we get


a + 6d – a – 11d = 34 – 64


⇒ – 5d = –30


⇒ d = 6


Putting the value of d in eq (1), we get


a + 6(6) = 34


⇒ a + 36 = 34


⇒ a = –2


Hence, the required AP is –2, 4, 10, 16,…


Now, we to find the 10th term


So, an = a + (n – 1)d


a10 = –2 + (10 – 1)6


a10 = –2 + 9 × 6


a10 = 52


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