Question

The sum of the first and 4th terms of arithmetic progression is 19 and the sum of the third and 6th term of A. P is 29 find ?

A) first term
B) common different

Answer 1

Answer:

$\sf{A) \ First \ term=5.75} \\ \\ \sf{B) \ Common \ difference =2.5}$

Given:

$\sf{\leadsto{Sum \ of \ first \ and \ forth \ term \ of \ A.P.=19}} \\ \\ \sf{Sum \ of \ third \ and \ sixth \ term \ of \ A.P.=29}$

To find:

$\sf{A) \ First \ term.} \\ \\ \sf{B) \ Common \ difference. }$

Solution:

$\boxed{\sf{t_{n}=a+(n-1)d}} \\ \\ \textsf{According to the first condition} \\ \\ \sf{t_{1}+t_{4}=19} \\ \\ \sf{\therefore{a+a+3d=19}} \\ \\ \sf{\therefore{2a+3d=19...(1)}} \\ \\ \textsf{According to the second condition} \\ \\ \sf{t_{3}+t_{6}=29} \\ \\ \sf{\therefore{t_{3}+t_{6}=29}} \\ \\ \sf{\therefore{a+2d+a+5d=29}} \\ \\ \sf{\therefore{2a+7d=29...(2)}} \\ \\ \textsf{Subtract equation (1) from equation (2), we get} \\ \\ \sf{2a+7d=29} \\ - \\ \sf{2a+3d=19} \\ \underline{\qquad} \\ \\ \sf{4d=10} \\ \\ \boxed{\sf{\therefore{d=2.5}}} \\ \\ \textsf{Substitute d=2.5 in equation (1), we get} \\ \\ \sf{2a+3(2.5)=19} \\ \\ \sf{\therefore{2a=19-7.5}} \\ \\ \sf{\therefore{2a=11.5}} \\ \\ \boxed{\sf{\therefore{a=5.75}}} \\ \\ \purple{\tt{A) \ First \ term=5.75}} \\ \\ \purple{\tt{B) Common \ difference=2.5}}$