Question

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. (Assume the four consecutive terms in A.P. are a - d , a, a + d, a + 2d.) Solve the word problem

Answer 1

Your answer is ---

Let, Four consevutive term in an AP be
(a-d),(a),(a+d)&(a+2d)

Now , according to given condition

a-d+a+a+d+a+2d = 12

=> 4a + d = 12 .....(1)

Since, sum of 3rd and 4th term is 14.

So,

a+d+a+2d = 14

=> 2a + 3d = 14 .... (2)

Multiply equation (2) by 2 , we get

4a + 12d = 56 ......(3)


Subtract equation (1) from (3) , we get

4a + 12d - (4a+d) = 56 - 12

=> 4a + 12d - 4a - d = 44

=> 11d = 44

=> d = 44/11

=> d = 4

put this value in equation (1), we get

4a + 4 = 12

=> a +1 = 12/4 = 3

=> a = 3 -1

=> a = 2

Therefore , a = 2 and b = 4

So, four consecutive term in an AP is
(2-4),(2),(2+4),(2+2×4)
= -2 , 2 , 6 , 10

Hence, AP: -2,2,6,10


【 Hope it helps you 】

Answer 2

Let a - 2d , a - d , a + d , a + 2d

are four consecutive terms in

A.P .

i )Sum of these terms = 12

=> a-2d+a-d+a+d+a+2d = 12

=> 4a = 12

=> a = 12/4

a = 3

ii ) Sum of 3rd and 4th term = 14

=> a+2d+a+3d = 14

=> 2a + 5d = 14

=> 2×3 + 5d = 14

=> 6 + 5d = 14

=> 5d = 14 - 6

=> 5d = 8

=> d = 8/5

Therefore ,

Four terms are ,

a - 2d = 3 - 2×8/5 = (15-16)/5=-1/5

a-d = 3 - 8/5 = (15-8)/5 = 7/5

a+d = 3+8/5 = (15+8)/5 = 23/5

a+2d=3+2×8/5 = (15+16)/5 = 31/5

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