Suppose that all the terms of an arithmetic progression (A.P) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

Sequences and Series PYQ — JEE Advanced 2015 · Shift 2 · Q27 with Solution

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Correct answer: 9

Let the A.P. have first term $a$ and common difference $d$ .
Sum of first $n$ terms of an A.P. is $S_n = \frac{n}{2}\left[2a + (n-1)d\right].$

So, $S_7 = \frac{7}{2}(2a+6d)=7(a+3d),$ $S_{11} = \frac{11}{2}(2a+10d)=11(a+5d).$

Given $S_7 : S_{11} = 6:11.$ Hence, $\frac{7(a+3d)}{11(a+5d)} = \frac{6}{11}.$ Cancel $11$ from both sides: $\frac{7(a+3d)}{a+5d} = 6.$ Thus, $7(a+3d)=6(a+5d).$ So, $7a+21d=6a+30d,$ $a=9d.$
The 7th term is $T_7 = a+6d = 9d+6d = 15d.$ Given that the 7th term lies between 130 and 140, $130<15d<140.$ Divide by 15: $\frac{130}{15}<d<\frac{140}{15},$ $8.\overline{6}<d<9.\overline{3}.$

Since all terms are natural numbers, $d$ must be a natural number. Therefore, $d=9.$

Check: $a=9d=81.$ Then the A.P. is $81,90,99,\dots$ The 7th term is $81+6\cdot 9=135,$ which is indeed between 130 and 140.

So the common difference is $\boxed{9}.$