Q10 of 47 Page 1

The 5th term of an A.P. is 26 and its 10th term is 51. Find the A.P.

Given: a5 = 26 and a10 = 51

We have to find the A.P.


We know that the nth term or the general term of an A.P is an = a + (n – 1) d where a = first term; n = number of terms and d = common difference.


So, a5 = a + (5 – 1) d


⇒ 26 = a + 4d ... (1)


Now, a10 = a + (10 – 1) d


⇒ 51 = a + 9d … (2)


Subtracting (1) from (2), we get


⇒ (a – a) + (9d – 4d) = 51 – 26


⇒ 5d = 25


∴ d = 5


Substituting value of d in (1)


⇒ a + 4(5) = 26


⇒ a = 26 – 20


∴ a = 6


We know that the general form of an A.P. is a, a + d, a + 2d, a + 3d…


So, 6, 6 + 5, 6+ 2(5), 6 + 3(5)…


⇒ 6, 11, 16, 21… is the required A.P.


Ans. The A.P. formed by the given conditions is 6, 11, 16, 21…


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