Question

Sum of10thand70thterm ofanarithmeticsequenceis806. Ifits20thterm is203,findits60thterm.

Answer 1

$\huge{\underline{\bf\red{Questi {\mathbb{O}} n} :}}$

$\sf Sum \: of \: {10}^{th} \: and \: {70}^{th} \: terms \: of \: an \: \\ \sf arithmetic \: progression \: is \: 806. \\ \sf If \: its \: {20}^{th} \: term \: is \: 203 , find \: its \\ \sf {60}^{th} \: term.$

$\huge{\underline{\: \bf\green{Answ {\mathbb{E}} r}\ \: :}}$

The answer is : $\boxed{\blue{{\bf\green{\quad 603 \quad}}}}$

$\Large{\underline{\bf\pink{Giv {\mathbb{E}} n}\ :}}$

• sum of $10^{ \sf th}$ and $70^{ \sf th}$ term = 806

• $20^{ \sf th}$ term = 203

$\Large{\underline{\bf\pink{To \ Fi {\mathbb{N}} d}\ :}}$

• $60^{ \sf th}$ term

$\huge{\underline{\bf\blue{Soluti {\mathbb{O}} n}\ :}}$

Let us assume that :-

• first term = a

• common difference = d

• nth term = $\sf{a_{n}}$

Now,

A/Q,

$\begin{aligned} \sf a_{10} + a_{70} & = 806 \\ \\ \implies \sf(a + 9d) + (a + 69d) & = 806 \\ \\ \implies \quad \ \ \sf a + a + 9d + 69d& = 806 \\ \\ \implies \sf \qquad \qquad \quad 2a + 78d & = 806 \\ \\ \implies \sf \qquad \qquad \ \cancel{2}(a + 39d) & = \cancel{2}(403) \\ \\ \implies \sf \qquad \qquad \quad \ \ \underline{\underline{a + 39d}} & = \underline{\underline{403}} - - - - - (i) \ \bigstar & & & & \end{aligned}$

And,

$\begin{aligned} \sf a_{20} & = 203 \\ \\ \implies \sf \underline{ \underline{a + 19d}} & = \underline{ \underline{203}} - - - - - (ii) \ \bigstar & & & & & & & & \end{aligned}$

On solving equations ( i ) and ( ii ) , we get :--

• first term , a = 13

• common difference , d = 10

..

..

( finding 60th term )

..

..

$\large \mathbb{HENCE ,}$

$\begin{aligned} \sf a_{60} & = \sf a + 59d \\ \\ & = (13) + 59(10) \\ \\ & = 13 + 590 \\ \\ & = \boxed{603} & & & & & \end{aligned}$

$\red{\rule{5.5cm}{0.02cm}}$

Formulae Used :

• $\sf a_n = a + (n - 1)d$

$\purple{\rule{7cm}{0.01cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$