The sum of first n terms of the A.P 3 , 5½ , 8 , … is equal to the 2nth term of the A.P 161/2
, 28 ½ , 40 ½ , …Find the value of n
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For the A.P. : 3, 5 1 2 512, 8, .... a = 3, d = 2 1 2 212,
number of terms = n ⇒ Sn = n 2 n2 [2a + (n – 1) d] =
n 2 n2 ( 6 + (n -1) × × 5 2 52) ... (i)
For the A.P.: 16 1 2 1612,
28 1 2 2812,
40 1 2 4012 ...
a = 16 1 2 1612,
d = 12. ∴ T2n= a + (2n – 1)
d = 16 1 2 1612 + (2n - 1) × ×12 ... (ii) Given, Sn = T2n ⇒ n 2 n2(6 +(n-1) 5 2 52) = 33 2 332+ (2n -1) × ×12 ...
[From (i) and (ii)] ⇒ Neglecting – n
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