Question

In an Ap the sum of the 1st 3terms is -36 and that of last 3 is 27. If there are 10 terms what is the 1st term and common difference?

Answer 1

Answer:

• Number of Terms (n) = 10

Let the First Term be a and Common Difference be d of AP.

$\underline{\bigstar\:\boldsymbol{Sum\:of\:First\:3\:Terms :}}$

$:\implies\sf T_1+T_2+T_3=-36\\\\\\:\implies\sf a+(a+d)+(a+2d)=-36\\\\\\:\implies\sf 3a+3d=-36\\\\\\:\implies\sf 3(a+d)=-36\\\\\\:\implies\sf a+d=-12\\\\\\:\implies\sf a=-12-d\qquad...(i)$

$\rule{120}{1}$

$\underline{\bigstar\:\boldsymbol{Sum\:of\:Last\:3\:Terms :}}$

$:\implies\sf T_8+T_9+T_{10}=27\\\\\\:\implies\sf (a+7d)+(a+8d)+(a+9d)=27\\\\\\:\implies\sf 3a+24d=27\\\\\\:\implies\sf 3(a+8d)=27\\\\\\:\implies\sf a+8d=9\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Putting the value of a from eq.(i)}}\\\\:\implies\sf -12-d+8d=9\\\\\\:\implies\sf -d+8d=9+12\\\\\\:\implies\sf 7d=21\\\\\\:\implies\underline{\boxed{\sf d=3}}$

⠀

$\therefore\:\underline{\textsf{Common Difference of the AP is \textbf{3}}}.$

$\rule{170}{2}$

$\underline{\bigstar\:\boldsymbol{Putting\:value\:of\:d\:in\:eq\:(i) :}}$

$:\implies\sf a=-12-d\\\\\\:\implies\sf a=-12-3\\\\\\:\implies\underline{\boxed{\sf a=-\:15}}$

⠀

$\therefore\:\underline{\textsf{First Term of the AP is \textbf{- 15}}}.$

Answer 2

Answer:

a=-15 (first term)

d = 3 (common difference)

Step-by-step explanation:

Given,

Sum of first three terms = -36

Sum of last three terms = 27

No.of.terms = 10

First term ,(a)= ?

Common difference (d)= ?

Let us suppose the terms from first,

Let terms be

a , a+d , a+2d

=>a+a+d+a+2d = -36

=> 3a+3d = -36

=> a + d = -12

=> a = -12-d

=> a = -12-3

=> a = -15

Let us suppose the terms from last be

a+9d,a+8d,a+7d

=> a+9d+a+8d+a+7d = 27

=> 3a+24d= 27

=> a+8d = 9

=> -12-d+8d= 9

=> 7d = 21

=> d = 3

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