Question

how many terms of am AP ., 1,5,9,............ are needed to make the sum 1225.

Please give fast answer and the answer is (25)​

Answer 1

Answer:

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Here a=1,d=5-1=4.

Let there are n terms which make the sum=1225

we know

Sn=n/2×(2a+(n-1)×d)=1225

1225×2=n(2a+(n-1)×d)

2450=n(2×1+(n-1)×4)

2450=n(2+4n-4)

2450=n(-2+4n)

2450=-2n+4n²

1225=-n+2n²

2n²-n-1225=0

Use factorisation to find the roots

2n²-50n+49n-1225=0

2n(n-25)+49(n-25) =0

(2n+49)(n-25) =0

either

2n+49=0. n-25=0

n≠-49/2. n=25

Therefore there are $\huge\blue{\boxed{\sf 25 \:terms}}$ In the AP

Answer 2

Answer:

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Step-by-step explanation:

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