Question

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 between 130 and 140, then the common difference of this A.P is
A) 13
B) 27
C) 9
D) 12

Answer 1

First of all calculate the sum of the first seven terms and first eleven terms using the formula of ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$. Then calculating their ratios and equating it with the given condition. Then write the seventh term of an A.P using $Ta + \left( {n - 1}\right)d$. After that from the above equation convert the equation of the seventh term of an A.P in terms of d and the given range of value of d, calculate for the natural values of d as given that all the terms are in the natural number of an A.P.

Complete step-by-step answer:
Given: - As the given ratios of the sum of the seventh and eleventh term are 6: 11.
So, write the sum up to the given terms and take their ratios using the formula,
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
The sum of the 7 th term is,
$\Rightarrow {S_7} = \dfrac{7}{2}\left[ {2a + \left( {7 - 1} \right)d} \right]$
Simplify the terms,
$\Rightarrow {S_7} = \dfrac{7}{2}\left[ {2a + 6d} \right]$
Take 2 commons from the bracket,
$\Rightarrow {S_7} = \dfrac{7}{2} \times 2\left( {a + 3d} \right)$
Cancel out the common terms and multiply the remaining values,
$\Rightarrow {S_7} = 7a + 21d$ ….. (1)
The sum of the 11 th term is,
$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {2a + \left( {11 - 1} \right)d} \right]$
Simplify the terms,
$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {2a + 10d} \right]$
Take 2 commons from the bracket,
$\Rightarrow {S_{11}} = \dfrac{{11}}{2} \times 2\left( {a + 5d} \right)$
Cancel out the common terms and multiply the remaining values,

$\Rightarrow {S_{11}} = 11a + 55d$ ….. (2)
As the given ratios of the sum of the seventh and eleventh term are 6: 11.
$\dfrac{{{S_7}}}{{{S_{11}}}} = \dfrac{6}{{11}}$
Substitute the values from equation (1) and (2),
$\Rightarrow \dfrac{{7a + 21d}}{{11a + 55d}} = \dfrac{6}{{11}}$
Cross-multiply the terms,
$\Rightarrow 77a + 231d = 66a + 330d$
Move $66a$ on the left side and $231d$ on the right side,
$\Rightarrow 77a - 66a = 330d - 231d$
Subtract the values on both sides,
$\Rightarrow 11a = 99d$
Divide both sides by 11,
$\therefore a = 9d$
Now, calculate the value of the seventh term,
${T_7} = a + \left( {7 - 1} \right)d$
Substitute the values,
${T_7} = 9d + 6d$
Add the terms on the right side,
${T_7} = 15d$
Since the term ranges between 135 and 140. So, the value of the seventh term should be a multiple of
15 between 135 and 140.
Then, 135 is the number that lies between the given range and satisfying the above condition so,
${T_7} = 135$
Substitute the values,
$15d = 135$
Divide both sides by 15,
$d = 9$

Hence, option (C) is the correct answer.

Note: In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. The difference here means the second minus the first. The is one more sum formula for the AP that is $\dfrac{n}{2}\left( {a + l} \right)$, where a is the first term and l is the last term of the series.