If the  ratio of the  sum  of n terms of two  AP’s be (7n  + 1):(4n + 27),  then the  ratio of their 11th  terms will be

If the  ratio of the  sum  of n terms of two  AP's be (7n  + 1):(4n + 27),  then the  ratio of their 11th  terms will be

1. A

2:3

2. B

3:4

3. C

4:3

4. D

5:6

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Solution:

Let Sn, and S' be the sums of n terms of two AP's and T11 and T11, be the respective 11th term, then

$\frac{{\mathrm{S}}_{\mathrm{n}}}{{\mathrm{S}}_{\mathrm{n}}^{\mathrm{\prime }}}=\frac{\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]}{\frac{\mathrm{n}}{2}\left[2{\mathrm{a}}^{\mathrm{\prime }}+\left(\mathrm{n}-1\right){\mathrm{d}}^{\mathrm{\prime }}\right]}=\frac{7\mathrm{n}+1}{4\mathrm{n}+27}$ (given)

$⇒$$\frac{\mathrm{a}+\frac{\left(\mathrm{n}-1\right)}{2}\mathrm{d}}{{\mathrm{a}}^{\mathrm{\prime }}+\frac{\left(\mathrm{n}-1\right)}{2}{\mathrm{d}}^{\mathrm{\prime }}}=\frac{7\mathrm{n}+1}{4\mathrm{n}+27}$

Now, put n=21, we get

$\frac{\mathrm{a}+10\mathrm{d}}{{\mathrm{a}}^{\mathrm{\prime }}+10{\mathrm{d}}^{\mathrm{\prime }}}=\frac{{\mathrm{T}}_{11}}{{\mathrm{T}}_{11}^{\mathrm{\prime }}}=\frac{148}{111}=\frac{4}{3}$

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