Question

14. Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their
squares is 140.

Answer 1

Hi ,

Let ( a - d ) , a , ( a + d ) are three consecutive

terms in an A.P

sum of the three terms = 18 ( given )

a - d + a + a + d = 18

3a = 18

a = 18/3

a = 6 ----( 1 )

sum of the squares of three terms = 140

( a - d )² + a² + ( a + d )² = 140

2( a² + d² ) + a² = 140

2a² + 2d² + a² = 140

3a² + 2d² = 140

3 × 6² + 2d² = 140 [ from ( 1 ) ]

108 + 2d² = 140

2d² = 140 - 108

2d² = 32

d² = 32/2

d² = 16

d = ± √16

d = ± 4

Therefore ,

Required three consecutive terms are

i ) if a = 6 , d = 4

a - d = 6 - 4 = 2

a = 6

a + d = 6 + 4 = 10

Or

ii ) if a = 6 , d = -4

10 , 6 , 2

I hope this helps you.

: )