Question

the ratio of the sum of n terms of an AP is 5(2n+1):97-2n. if the first term and common difference of the first AP is 3 and 4. Find the second AP. [ans-19, 18whole1/5, 17, 16 whole 3/5​

Answer 1

★$\huge\bf{\blue{\underline{Question:-}}}$

the ratio of the sum of n terms of an AP is 5(2n+1):97-2n. if the first term and common difference of the first AP is 3 and 4. Find the second AP. [ans-19, 18whole1/5, 17, 16 whole 3/5

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★$\huge\bf{\red{\underline{Answer:-}}}$

the ratio of the mth terms of two AP's is (14m−6):(8m+23)

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★$\huge\bf{\green{\underline{Explanation:-}}}$

Let a1, a2 be the first terms and d1, d2 the common differences of the two given A.P.'s. Then, the sums of their n terms are given by

Sn=2n[2a1+(n−1)d1]

And,

Sn′=2n[2a2+(n−1)d2]

Therefore,

Sn′Sn=2n[2a2+(n−1)d2]2n[2a1+(n−1)d1]=2a2+(n−1)d22a1+(n−1)d1

It is given that

Sn′Sn=4n+277n+1

⟹2a2+(n−1)d22a1+(n−1)d1=4n+277n+1     ....(1)

To find the ratio of the mth terms of the two given AP's, we replace n by (2m-1) in equation 1.

Therefore,

2a2+(2m−2)d22a1+(2m−2)d1=4(2m−1)+277(2m−1)+1

⟹a2+(m−1)d2a1+(m−1)d1=8m+2314m−6

Hence, the ratio of the mth terms of two AP's is (14m−6):(8m+23).

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Answer 2

$\huge \tt \pink {answer..}$

Let a1, a2 be the first terms and d1, d2 the common differences of the two given A.P.'s. Then, the sums of their n terms are given by

Sn=2n[2a1+(n−1)d1]

And,

Sn′=2n[2a2+(n−1)d2]

Therefore,

Sn′Sn=2n[2a2+(n−1)d2]2n[2a1+(n−1)d1]=2a2+(n−1)d22a1+(n−1)d1

It is given that

Sn′Sn=4n+277n+1

⟹2a2+(n−1)d22a1+(n−1)d1=4n+277n+1 ....(1)

To find the ratio of the mth terms of the two given AP's, we replace n by (2m-1) in equation 1.

Therefore,

2a2+(2m−2)d22a1+(2m−2)d1=4(2m−1)+277(2m−1)+1

⟹a2+(m−1)d2a1+(m−1)d1=8m+2314m−6

Hence, the ratio of the mth terms of two AP's is (14m−6):(8m+23).

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