JEE Advanced 2015 · Paper 2 · Mathematics Q42

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MathematicsInteger+4Question 42 of 60

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

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Answer and worked solution

Official answer: 9

Ratio of sums.
$\frac{S_7}{S_{11}}=\frac{7(a+3d)}{11(a+5d)}=\frac6{11}\implies a=9d.$

Seventh term. $a_7=a+6d=15d$ lies between 130 and 140. With $d$ a natural number, $15d=135$ , so $d=9$ .

Answer: 9

Topics:Arithmetic progression

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JEE Advanced 2015 · Paper 2, Mathematics question 42