Question

Which of the following statements is INCORRECT?

• A. If $S _1, S_2, S_3$ are the sum of first $n$ natural numbers, their squares and their cubes, respectively, then $7S_{2}^{2} = S_{3}(1 + 8S_{1})$.
• B. A farmer buys a used tractor for $₹\, 12000$. He pays $₹\,6000$ cash and agrees to pay the balance in annual installments of $₹\, 500$ plus $12\% $ interest on the unpaid amount. The tractor will cost him for $₹\, 16680$.
• C. If the $p^{th }$ and $q^{th}$ terms of a $G.P$. are $q$ and $p$ respectively, then its $\left(p + q\right)^{th}$ term is $\left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}$.
• D. None of these

Answer 1

Answer: (A)

(a) If $S_{1},S_{2,} S_{3}$ are the sum of first $n$ natural numbers, their squares and their cubes respectively, then
$S_{1} = \sum\limits_{k=1}^{n} k= \frac{n\left(n+1\right)}{2}$,
$ S_{2} = \sum\limits _{k=1}^{n} k^{2} = \frac{n\left(n+1\right)\left(2n+1\right)}{6} $
$ S_{3} = \sum \limits_{k=1}^{n} k^{3} \left[\frac{n\left(n+1\right)}{2}\right]^{2}$
Now, $1+8S_{1} = 1+8\left[\frac{n\left(n+1\right)}{2}\right]$
$ = 1+4n\left(n+1\right) = \left(2n+1\right)^{2} $
$ \therefore S_{3} \left(1+8S_{1}\right) = \left[ \frac{n\left(n+1\right)}{2}\right]^{2} \left(2n+1\right)^{2} $
$= \left[\frac{n\left(n+1\right)\left(2n+1\right)}{2}\right]^{2}$
$ = \frac{9}{9} \left[\frac{n\left(n+1\right)\left(2n+1\right)}{2}\right]^{2}$
$ =9\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]^{2} = 9S_{2}^{2}$
(b) Farmer purchased a tractor for $₹ \,12,000$. After paying $₹ \,6000$, he will left with $₹\, 6000$.
$ 1^{st}$ interest paid $= ₹\, \frac{6000\times12}{100} = ₹\,720$
$ 2^{nd }$ interest paid $= ₹ \,\frac{5500\times12}{100} = ₹\, 660$
$⋮ \qquad ⋮ \qquad ⋮$
$12^{th} $ interest paid $= ₹\, \frac{500\times12}{100} = ₹ \,60$
$ \therefore $ Total amount of interest $= ₹\, \left[720+660+.....+60\right]$
$=₹\,\left[\frac{12}{2}\left(720+60\right)\right] = ₹\, \left(6\times780\right)= ₹\,4680$
$\therefore $ The cost of tractor $= ₹\, \left(12000+4680\right) = ₹ \,16680$.
$ \left(c\right) T_{p}= q$
$ \Rightarrow ar^{p-1} = q\quad...\left(i\right)$
and $T_{q} =p $
$ \Rightarrow ar^{q-1}= p \quad...\left(ii\right)$
On dividing $\left(i\right)$ by $\left(ii\right)$, we get
$ \frac{ar^{p-1}}{ar^{q-1}} = \frac{q}{p} $
$\Rightarrow r^{p-q} = \frac{q}{p}$
$\Rightarrow r= \left(\frac{q}{p}\right)^{\frac{1}{p-q}} \quad ...\left(iii\right)$
Putting the value of $r$ in $\left(i\right)$, we get
$ a\left(\frac{q}{p} \right)^{\frac{p-1}{p-q}} = q $
$ \Rightarrow a = \frac{q}{ \left(\frac{q}{p}\right)^{\frac{p-1}{q-1}}} = q\cdot\left(\frac{p}{q}\right) ^{\frac{p-1}{p-q}} \quad ...\left(iv\right)$
$ \therefore \left(p+q\right)^{th }$ term, $T_{p+q} = a\cdot r^{p+q-1}$
$= q\cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}}\left[\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\right]^{p+q-1}$ [using $(iii)$ and $(iv)$]
$= q\cdot\left(\frac{p}{q}\right)^{\frac{p-1}{p-q}} \left(\frac{q}{p}\right)^{\frac{p+q-1}{p-q}} = q\cdot \left(\frac{p}{q}\right)^{\frac{p-1}{p-q}-\frac{\left(p+q-1\right)}{p-q}}$
$ = q\cdot\left(\frac{p}{q}\right)^{\frac{-q}{p-q}} = q\left(\frac{q}{p}\right)^{\frac{q}{p-q}}= \frac{q^{\frac{q}{p-q}+1}}{p^{\frac{q}{p-q}}}$
$ = \frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}} = \left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}$