Q31 of 40 Page 1

Find a, b and c if it is given that the numbers a, 7, b, 23, c are in A.P.]

First term of given AP is ‘a’, let common difference be ‘d’.


OR


If m times the mth term of an AP is equal to n times its nth term, show that the (m+n)th term of the AP is zero.


We have,

Second term, a2 = 7


⇒ a + d = 7 [1]


Fourth term, a4 = 23


⇒ a + 3d = 23 [2]


[2] – [1] gives,


3d – d = 23 – 7


⇒ 2d = 16


⇒ d = 8


⇒ a + 8 = 7


⇒ a = -1


b = 7 + d = 7 + 8 = 15


c = 23 + d = 23 + 8 = 31


OR


We know that nth term of A.P. , tn = a + (n – 1) d.


First, tn = a + (n – 1) d


Then, tm = a + (m – 1) d


Given, mtm = ntn


⇒ m [a + (m – 1) d] = n [a + (n – 1) d]


⇒ ma + m2d – md = na + n2d – nd


⇒ ma – na + m2d – n2d – md + nd = 0


⇒ a (m – n) + d (m2 – n2) – d (m – n) = 0


We know that a2 – b2 = (a – b) (a + b)


⇒ (m – n) [a + d (m + n) – d] = 0


⇒ [a + d (m + n) – d] = 0


⇒ a + (m + n – 1) d = 0


∴ (m + n)th term, tm + n = 0


Hence proved.


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