Question

Find the A.P whose first term is 100 and the sum of first six terms is 5 times the sum of next 6 terms

Answer 1

Answer:

Here a=100

Let difference is d.

⇒a1+a2+a3+a4+a5+a6 = 5(a7+a8+a9+a10+a11+a12)

So by the formula, Sn = n/2(a+l), where a&l are the first and last term of an AP, we have

6(2a1+a6) = 5×6(2a7+a12)

⇒a1 + a6 = 5(a7+a12)

⇒a+a+5d = 5(a+6d+a+11d)

⇒2a+5d = 10a+85

⇒80d = −8a

⇒d = −a/10

d = −100/10

d = −10

Therefore, 100,90,80,70,........is an required A.P

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