Question

the sum of three consecutive terms of an increasing A.P. is 42 and the product of the first and third of these terms is 187, then the third term is *
1)11
2)17
3)21
4)14

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Answer 1

Answer:

11,17

Step-by-step explanation:

let the numbers in AP are a-d,a,a+d

given sum of them is 42

a-d+a+a+d=42

3a=42

a=14

product of first and third term is 187

(a-d)(a+d)=187

(14-d)(14+d)=187

$14^{2}-d^{2} =187\\14^{2} -187=d^{2}$

196-187=$d^{2}$

9=$d^{2}$

d=3,-3

the numbers are 11,14,17

or 17,14,11

Answer 2

let the numbers in AP are a-d,a,a+d

given sum of them is 42

a-d+a+a+d=42

3a=42

a=14

product of first and third term is 187

(a-d)(a+d)=187

(14-d)(14+d)=187

196-187= d²

9=d²

d=3,-3

the numbers are 11,14,17

or 17,14,11..