Question

The ratio of the sums of m and n terms of an A.P is $\bf{m^2:n^2}$ Show that the ratio of mth and nth term is (2m−1):(2n−1).

(No spamming $\checkmark$)​

Answer 1

★ Concept :-

Here the concept of Sum of n terms of an A.P. and nth term of an A.P. have been used. We see that we are given the ratio of sum of first m and sum of first n terms of an A.P. From this we can find different values like first term and common difference of the A.P. Also we can find certain relationships. After that we can apply another formula to find the mth and nth term of the A.P. and after that we can find the required ratio.

Let's do it !!

______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{S_{p}\;=\;\dfrac{p}{2}[2a\:+\:(p\:-\:1)d]}}}$

$\;\boxed{\sf{\pink{a_{p}\;=\;a\:+\:(p\:-\:1)d}}}$

______________________________________

★ Solution :

Given,

» Sum of first m terms of an A.P. : Sum of first n terms of the A.P. = m² : n²

This means that for both m and n terms, A.P. is same. This also means that First term and Common difference for both will be same.

• Let the common difference of the A.P. be d

• Let the first term of the A.P. be a

» Sum of first m terms = Sm

» Sum of first n terms = Sn

We know the formula that,

$\;\sf{\rightarrow\;\;S_{p}\;=\;\dfrac{p}{2}[2a\:+\:(p\:-\:1)d]}$

• Here Sp is the sum of first p terms of an A.P.

Let's now find the Sm and Sn using this formula.

• For the value of Sm ::

Here since we need to find the sum of first m terms, we need to replace p with m. So,

$\;\bf{\rightarrow\;\;\green{S_{m}\;=\;\dfrac{m}{2}[2a\:+\:(m\:-\:1)d]}}$

• For the value of Sn ::

Here since we need to find the sum of first n terms, we need to replace p with n. So,

$\;\bf{\rightarrow\;\;\blue{S_{n}\;=\;\dfrac{n}{2}[2a\:+\:(n\:-\:1)d]}}$

• For the different relationships ::

According to the qúestion,

$\;\tt{\rightarrowtail\;\;\dfrac{\green{S_{m}}}{\blue{S_{n}}}\;=\;\dfrac{m^{2}}{n^{2}}}$

Now by applying the values, we get

$\;\tt{\rightarrowtail\;\;\dfrac{\green{\bigg(\dfrac{m}{2}[2a\:+\:(m\:-\:1)d]\bigg)}}{\blue{\bigg(\dfrac{n}{2}[2a\:+\:(n\:-\:1)d]\bigg)}}\;=\;\dfrac{m^{2}}{n^{2}}}$

On cancelling 2 from both numerator and denominator, we get

$\;\tt{\rightarrowtail\;\;\dfrac{m[2a\:+\:(m\:-\:1)d]}{n[2a\:+\:(n\:-\:1)d]}\;=\;\dfrac{m^{2}}{n^{2}}}$

Now on taking the like terms to opposite side, we get

$\;\tt{\rightarrowtail\;\;\dfrac{[2a\:+\:(m\:-\:1)d]}{[2a\:+\:(n\:-\:1)d]}\;=\;\dfrac{m^{2}\:\times\:n}{n^{2}\:\times\:m}}$

On cancelling these like terms, we get

$\;\tt{\rightarrowtail\;\;\dfrac{[2a\:+\:(m\:-\:1)d]}{[2a\:+\:(n\:-\:1)d]}\;=\;\dfrac{m}{n}}$

Now by cross multiplication of L.H.S. and R.H.S., we get

>> n[2a + (m - 1)d] = m[2a + (n - 1)d]

On further multiplication, we get

>> n[2a + dm - d] = m[2a + dn - d]

>> 2an + dmn - dn = 2am + dnm - dm

On transposing each term from R.H.S. to L.H.S., we get

>> 2an + dmn - dn - 2am - dnm + dm = 0

>> 2an - dn - 2am + dm = 0

>> 2an - 2am - dn + dm = 0

On taking common terms, we get

>> 2a(n - m) + d(m - n) = 0

In order to do grouping of common terms, we need same terms to be multiplied. So, we have to change the sign here. Then,

>> 2a(n - m) - d(n - m) = 0

On grouping,

>> (2a - d)(n - m) = 0

Here either (2a - d) = 0 or (n - m) = 0. So,

>> 2a - d = 0 or n - m = 0

>> 2a = d or n = m

This is the required relationship.

We know that formula that,

$\;\sf{\Longrightarrow\;\;a_{p}\;=\;a\:+\:(p\:-\:1)d}$

• Here ap is the pth term.

• For the mth term ::

In order to find the mth term, we simply need to replace p from the formula with m since other values are same.

$\;\bf{\Longrightarrow\;\;\orange{a_{m}\;=\;a\:+\:(m\:-\:1)d}}$

• For the nth term ::

Similar like above, in order to find nth term, we need to replace p in the formula with n to find the value.

$\;\bf{\Longrightarrow\;\;\red{a_{n}\;=\;a\:+\:(n\:-\:1)d}}$

• For proving the required ratio ::

Here we need to find the ratio of mth term to nth term. So,

$\;\tt{\mapsto\;\;\dfrac{\orange{a_{m}}}{\red{a_{n}}}\;=\;\dfrac{\orange{a\:+\:(m\:-\:1)d}}{\red{a\:+\:(n\:-\:1)d}}}$

$\;\tt{\mapsto\;\;\dfrac{a_{m}}{a_{n}}\;=\;\dfrac{a\:+\:dm\:-\:d}{a\:+\:dn\:-\:d}}$

We know that, d = 2a (from above relation). Then,

$\;\tt{\mapsto\;\;\dfrac{a_{m}}{a_{n}}\;=\;\dfrac{a\:+\:2am\:-\:2a}{a\:+\:2an\:-\:2a}}$

$\;\tt{\mapsto\;\;\dfrac{a_{m}}{a_{n}}\;=\;\dfrac{2am\:-\:a}{2an\:-\:a}}$

On taking a common from both numerator and denominator, we get

$\;\tt{\mapsto\;\;\dfrac{a_{m}}{a_{n}}\;=\;\dfrac{a(2m\:-\:1)}{a(2n\:-\:1)}}$

On cancelling a, we get

$\;\boxed{\bf{\odot\;\;\purple{\dfrac{a_{m}}{a_{n}}\;=\;\dfrac{(2m\:-\:1)}{(2n\:-\:1)}}}}$

This is the required thing to be proved.

Hence, proved.

________________________________

★ More to know :-

$\;\sf{\leadsto\;\;Arithmetic\;Mean\;=\;\dfrac{a_{1}\;+\;a_{2}\;+\;a_{3}\;+\;...\;+\;a_{n}}{n}}$

Answer 2

★$\underline{\bf{We\: know\: that,}}$★

sum of x terms of an AP is given by, $\sf Sx=x/2(2a+(x-1)$

$\underline{\bf{Now\: coming\: to\: the \:question,}}$

Sum of m terms of AP$\sf=m/2(2a+(m-1)$

Sum of n terms of AP$\sf=n/2(2a+(n-1)$

$\underline{\bf{Given\: that,}}$

Sm:Sn$\sf=m^2:n^2$

$\underline{\bf{So,}}$

$\sf[m/2(2a+(m-1)d)] /[n/2(2a+(n-1)d)] =m^2/n^2$

$\sf[2a+(m-1)d]/[2a+(n-1)d]=m/n$

$\sf(2a+md-d)/(2a+nd-d)=m/n$

$\sf2an+mnd-nd=2am+mnd-md$

$\sf2an-nd=2am-md$

$\sf2an-2am=nd-md$

$\sf2a(n-m)=d(n-m)$

★$\boxed{\bf{2a=d - - - - - - -eq (1)}}$★

We know that, xth term of an AP is given by, ax=a+(n-1) d

So, the mth term of AP is$\sf am=a+(m-1)d$

nth terms of AP is$\sf an=a+(n-1)d$

So, the ratio of mth and nth term of AP is

$\sf am/an=[a+(m-1)d] /[2a+(n-1)d]$

$\sf=(a+(m-1)2a)/(a+(n-1)2a)$(from eq1)

$\sf=(a+2am-2a)/(a+2an-2a)$

$\sf=(2am-a)/(2an-a)$

$\sf=a(2m-1)/a(2n-1)$

$\sf=(2m-1)/(2n-1)$

★So★

$\sf am:an=2m-1:2n-1$

$\sf\color{darkviolet}Hence\: proved$

━━━━━━━━━━━━━━━━━━━━━━━━━━

Thankyou :)

$\looparrowright$$\sf\color{skyblue}You \:can\: also\: refer\: to\: the\: attachment$