Question

》the ratio of the sum of n terms of 2 APs is 7n+1:4n+27.
find the ratio of their 11th terms.
plz solve

Answer 1

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Answer 2

Let the two terms have

first term =a1 , b1

common differences = d1 , d2

So sum of n terms of first Ap is..........

S1= n/2[2a1 + (n-1)d1]

sum of n terms of second AP is......

S2 = n/2[ 2b1 + (n-1)d2]

According to question....The ratio....

S1/S2 = (7n+1)/(4n+27)

$\frac{\frac{n}{2}(2a1 +(n-1)d1)}{n/2(2b1 + (n-1)d2} = \frac{7n+1}{4n+27}$

[2a1 + (n-1)d1]/{2b1 + (n-1)d2} = {7n+1}/{4n+27}

dividing denominator and by 2 in first ratio

$\frac{a1 +\frac{(n-1)d1}{2}}{b1 + \frac{(n-1)d2}{2}} = \frac{7n+1}{4n+27}$.....................(1)

Now the ratio of 11th terms of two ap`s is

a11/b11 = [ a1 + (11-1)d1]/[b1 + (11-1)d2]

a11/b11 = [a1 +10d1]/{b1 + 10d2}

on multiplying by 2 in second ratio...

a11/b11 = [2a1 + 20d1]/ [2b1 + 20d1]

a11/b11 = sum of 21 terms of first/ sum of 21 terms of second Ap. (20 =21-1 = m-1)

Now according to (1)

a11/b11 =[7(21) +1]/[4(21) +27]

a11/b11=148/111

so ratio of 11th term=148/111