Question

the sum of first thirty terms of an A.P is equal to square of sum of first six terms of same A.P. Then show that 10a+145d=12a^2+60ad+75d^2​

Answer 1

Answer:-

Given:

Sum of first thirty terms of an AP = Square of sum of first six terms.

S(30) = [S(6)] * [S(6)]

The first six terms of an AP are a , a + d , a + 2d , a + 3d , a + 4d , a + 5d.

We know that,

S(n) = n/2 * [ 2a + (n - 1)d ]

S(30) = 30/2 * [ 2a + (30 - 1)d ]

S(30) = 15[ 2a + 29d]

S(30) = 30a + 435d

S(6) = 6/2 * [ 2a + (6 - 1)d ]

S(6) = 3[ 2a + 5d ]

S(6) = 6a + 15d

S(30) = [ S(6) ]²

=>30a + 435d = (6a + 15d)²

=> 30a + 435d = (6a)² + (15d)² + 2(6a)(15d)

[ (a + b)² = a² + b² + 2ab ]

=> 30a + 435d = 36a² + 225d² + 180ad

=> 3(10a + 145d) = 3(12a² + 75d² + 60ad)

(3 is being cancelled out)

=> 10a + 145d = 12a² + 60ad + 75d²

Hence, proved.