Question

Suppose that all the terms of an arithmetic progression (A.P) are natural numbers. If the ratio of the sum of first seven terms to the sum of first elven terms is ${\text{6:11}}$ and the seven-term lies between ${\text{130}}$ and ${\text{140}}$, then the common difference of this A.P is

Answer 1

First of all calculate the sum of first seven terms and first eleven terms using the formula of ${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}}$
And then calculating their ratios and equating it with the given condition. Then write the seventh term of an A.P using
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ and hence from the above equation convert the equation of the seventh term of an A.P in terms of d and in 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 solution: As the given ratios of the sum of the seventh and eleventh term are ${\text{6:11}}$.
So, write the sum up to the given terms and take their ratios,
Using ${{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}}$, we get,
$\dfrac{{{{\text{S}}_{\text{7}}}}}{{{{\text{S}}_{{\text{11}}}}}}{\text{ = }}\dfrac{{\dfrac{{\text{7}}}{{\text{2}}}{\text{(2a + (6)d)}}}}{{\dfrac{{{\text{11}}}}{{\text{2}}}{\text{(2a + (10)d)}}}}{\text{ = }}\dfrac{{\text{6}}}{{{\text{11}}}}$
Now, simplify the equation and convert the term in the form of a and d,

$$\Rightarrow \dfrac{{{\text{7(a + 3d)}}}}{{{\text{11(a + 5d)}}}}{\text{ = }}\dfrac{{\text{6}}}{{{\text{11}}}} \\\ \Rightarrow \dfrac{{{\text{7(a + 3d)}}}}{{{\text{(a + 5d)}}}}{\text{ = 6}} \\\$$

Now, cross multiply the denominator and simplify them,
$\Rightarrow {\text{7(a + 3d) = 6(a + 5d)}}$
On opening the bracket and simplifying we get,
$\Rightarrow {\text{7a + 21d = 6a + 30d}}$
On solving for like terms we get,
$\Rightarrow {\text{a = 9d}}$
Hence, now calculating the value of the seventh term as,

$${{\text{T}}_{\text{n}}}{\text{ = a + (7 - 1)d}} \\\ {\text{ = a + 6d}} \\\$$

As now put the value of a in terms of d as ${\text{a = 9d}}$,

$${\text{ = 9d + 6d}} \\\ {\text{ = 15d}} \\\$$

So, as given in the question the terms ranges between ${\text{130}}$ and ${\text{140}}$
Hence the value of seventh term should be multiple of ${\text{15}}$ between ${\text{130}}$ and ${\text{140}}$
As ${\text{135}}$ is the number that lies between the given range and satisfying the above condition so,

$${{\text{T}}_7}{\text{ = 15d = 135}} \\\ {\text{d = 9}} \\\$$

Hence, we can conclude that the value of d is ${\text{9}}$.

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.