Question

How many terms of the AP 63, 60, 57, 54,.... must be taken so that their sum is 693? Explain the double answer.

Answer 1

Given series,

63, 60 , 57 , 54.........

First term = 63

common difference = -3

Let that n terms of AP must be taken so that their sum will be 693

Sn = n/2[2a +(n-1)d]

On putting , a = 63 and d = -3 , solving for n,

We get

n= 21
n=22

In this case we get two values of n because here common difference d is negative. Here sum of 21 term is equal to sum of. terms of 22. Probably 22th will be zero, therefore we have two acceptable answers.

Answer 2

$\huge{hey}$

$\huge{ \mathfrak{here \: is \: answer}}$

$\bf{given}$

$\sf{a \tiny{1} = \large{63}}$

$\sf{a \tiny{2} = \large{60}}$
$\sf{a \tiny{3} = \large{57}}$

$\sf{a \tiny{4} = \large{54}}$

$\sf{d = \large{a }\tiny{2} - \large{a} \tiny{1}}$
=> 57-60

= -3

$\sf{ \large{s} \tiny{n} = \large{ \frac{n}{2} }(2a + (n - 1)d)}$

then we will solve on the basis of this formula

$\bf{see \: in \: pic}$

$\bf{hence \: answer \: is \: 21/: and/:22}$


$\huge \boxed{ \ulcorner{hope \: it \: helps}}$

$\huge{ \mathfrak{THANKS}}$