Question

find four consecutive term in an a.p. whose sum is 88 and the sum of the 1st and the 3rd term is 40

Answer 1

Let four consecutive terms are a - 3d , a - d , a + d , a + 3d , where a and d is real numbers.

A/C to question,
sum of all four terms = 88
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 88
4a = 88
a = 22 ......(1)

again, sum of 1st and 3rd term is 40
e.g., (a - 3d) + (a + d) = 40
2a - 2d = 40
a - d = 20
from equation (1),
22 - d = 20
d = 2

hence, a = 22 and d = 2
then, a - 3d = 22 - 3 × 2 = 22 - 6 = 16
a - d = 22 - 2 = 20
a + d = 22 + 2 = 24
a + 3d = 22 + 3× 2 = 22 + 6 = 28

therefore, four consecutive term in an A.P are 16, 20, 24, 28

Answer 2

Solution :

Let a , d are first term and common

difference of an A.P

Now ,

a , a+d , a+2d, a + 3d are first four

consecutive terms in A.P

Sum of 4 terms = 88 ( given )

=> a+a+d+a+2d+a+3d =88

=> 4a+6d = 88

Divide each term with 2 , we get

=> 2a + 3d = 44 ----( 1 )

Sum of first and 3rd term = 40(given )

=> a + a + 2d = 40

=> 2a + 2d = 40 ---( 2 )

subtract ( 2 ) from ( 1 ) , we get

d = 4

substitute d = 4 in equation (2),

we get

2a + 2×4 = 40

=> 2a = 40 - 8

=> 2a = 32

=> a = 32/2

=> a = 16

Therefore ,

a = 16 , d = 4

Required 4 terms are ,

a = 16 ,

a + d = 16 + 4 = 20 ,

a+2d = 16+2×4 = 24 ,

a+3d = 16+3×4 = 28

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