The question is given from ncert Book of class 10th Chapter no. 5 Ex. 5.2 Q. 18. In the following question you have to find the 1st three terms of an A.P in which the sum of 4th & 8th term is 24 and sum of 6th & 10th term is 44. Give the solution.

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Solution:The nth term of the AP is;

a=_{n}a+(n−1)da_{4}=a+(4−1)da_{4}=a+3dLike above, we can write,

a_{8}=a+7da_{6}=a+5da_{10}=a+9dIt is given that,

a

_{4}+a_{8}= 24a+3d+a+7d = 24

2a+10d = 24

a+5d = 12 c…………………………………………

(i)a

_{6}+a_{10}= 44a +5d+a+9d = 44

2a+14d = 44

a+7d = 22 …………………………..

(ii)By subtracting eq.

(i)from(ii),2d = 22 − 12

2d = 10

d= 5From eq.

(i), we get,a+5d= 12a+5(5) = 12a+25 = 12a= −13a_{2}=a+d= − 13+5 = −8a_{3}=a_{2}+d= − 8+5 = −3So, the 1st three terms of the Arithmetic Progression are −13, −8, and −3.