Question

(ii) Find five numbers in A.P., whose gum is 25 and the sum of whose squares is
135.

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Answer 1

Answer:

term of an ap is 3,4,5,6,7

Step-by-step explanation:

let the five term be (a-2d), (a-d) ,a (a+d), (a+2d)

then, sum of term is 25

(a-2d) + (a-d) + a (a+d) + (a+2d) = 25

5a = 25

a = 5

given , sum of squares is 135

(a-2d)^2 + (a-d)^2 + a^2 + (a+d)^2 + (a+2d)^2 = 135

put a=5 and evaluate, we get

4d^2 + d^2 + d^2 + 4d^2 = 135-125

10d^2 = 10

d^2 = 1

d = + - 1

when a = 5 and d = 1

(a-2d) = 5-2 = 3

(a-d) = 5-1 = 4

a = 5

(a+d) = 5+1 = 6

(a+2d) = 5+2 = 7

when a = 5 and d = -1

( a-2d) = 5+2 = 7

(a-d) = 5+1 = 6

a = 5

(a+d) = 5-1 = 4

(a+2d) = 5-2 = 3

thus , term of AP are 3,4,5,6 and 7