Question

The sum of the 4th and the 8th terms of an
A.P. is 24 and the sum of the 6th and the 10th
terms of the same A.P. is 34. Find the first
three terms of the A.P.​

Answer 1

Answer:

Step-by-step explanation:

i hope it's help you

Answer 2

$\sf\large\underline\blue{Given:}$

$\sf{\implies Sum\:_{(4th+8th)}=24}$

$\sf{\implies Sum\:_{(6th+10th)}=34}$

$\sf\large\underline\blue{To\: Find:}$

$\sf{\implies The\: first\:3\: terms\:AP=?}$

$\sf\large\underline\blue{Solution:}$

• At first calculate the value of a and d as per the given clue in the question for it we have to set up equation then solve the equation.]

$\sf{\implies Given\:in\:case\:(I)}$

• The sum of the 4th and the 8th terms of an A.P. is 24.

$\sf{\implies a_{(n)}=a+(n-1)d}$

$\sf{\implies a+3d+a+7d=24}$

$\sf{\implies 2a+10d=24}$

• Dividing by 2 on both sides:]

$\sf{\implies a+5d=12-----(i)}$

$\sf{\implies Given\:in\:case\:(ii)}$

• The sum of the 6th and the 10th terms of an A.P. is 34.]

$\sf{\implies a_{(n)}=a+(n-1)d}$

$\sf{\implies a+5d+a+9d=34}$

$\sf{\implies 2a+14d=34}$

• Dividing by 2 on both sides:]

$\sf{\implies a+7d=17------(ii)}$

• Eq (i) and (ii) solving here:]

$\sf{\implies a+5d=12}$

$\sf{\implies a+7d=17}$

• By solving we get, here]

$\sf{\implies -2d=-5\implies d=\dfrac{5}{2}}$

• Putting the value of d=5/2 in eq (i)

$\sf{\implies a+5d=12}$

$\sf{\implies a+\dfrac{25}{2}=12}$

$\sf{\implies a=12-\dfrac{25}{2}}$

$\sf{\implies a=\dfrac{-1}{2}}$

• Now calculate first 3 term here:]
• Let the terms be a+d, a , a-d

$\sf\large{Hence,}$

$\sf{\implies First\:_{(term)}=a+d}$

$\sf{\implies First\:_{(term)}=\dfrac{-1}{2}+\dfrac{5}{2}}$

$\sf{\implies First\:_{(term)}=\dfrac{4}{2}=2}$

$\sf{\implies Second\:_{(term)}=a}$

$\sf{\implies Second\:_{(term)}=\dfrac{-1}{2}}$

$\sf{\implies Third\:_{(term)}=a-d}$

$\sf{\implies Third\:_{(term)}=\dfrac{-1}{2}-\dfrac{5}{2}}$

$\sf{\implies Third\:_{(term)}=\dfrac{-6}{2}=-3}$