27. If the ratio of the sum of the first n terms of two A.Ps is (7n +

1.	27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their9terms.
Answer» Sn/sn =7n+1/4n+27 (given) Where Sn is sum of n terms of 1st AP and sn is sum of n terms of second AP. Sn =n/2(2A +(n-1)D) (A =first term, D =common difference) (Sum of AP formula) sn=n/2(2a+(n-1)d) (a=first term, d=common difference) (Sum of AP formula) Sn/sn=(2A+(n-1)D)/(2a+(n-1)d) 7n+1 /4n+27 = (2A+(n-1)D) / (2a+(n-1)d) As, we have to find ratio of their 9th term, let n =2(9)-1=17 So, equation=> 7(17)+1/4(17)+27=2A+(16D)/2a+16d 120/95 =A+8D /a+8d So, T9/t9 =120/95=24/19 Where T9 is 9th term of first AP and t9 is 9th term of second AP. Hit like if you find it useful!

27. If the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their9terms.

Answer»

Sn/sn =7n+1/4n+27 (given)

Where Sn is sum of n terms of 1st AP and sn is sum of n terms of second AP.

Sn =n/2(2A +(n-1)D) (A =first term, D =common difference) (Sum of AP formula)

sn=n/2(2a+(n-1)d) (a=first term, d=common difference) (Sum of AP formula)

Sn/sn=(2A+(n-1)D)/(2a+(n-1)d)

7n+1 /4n+27 = (2A+(n-1)D) / (2a+(n-1)d)

As, we have to find ratio of their 9th term, let n =2(9)-1=17

So, equation=>

7(17)+1/4(17)+27=2A+(16D)/2a+16d

120/95 =A+8D /a+8d

So, T9/t9 =120/95=24/19

Where T9 is 9th term of first AP and t9 is 9th term of second AP.

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