An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP

Answer

The required A.P. is `3, 7, 11, 15, …, 147.`

Explanation

Here, `n=37`
The middle term of the A.P.
`=(n+1)/(2)=(37+1)/(2)=(38)/(2)=19.`
`therefore` the middle terms are 18th, 19th and 20th.
`t_(18)+t_(19)+t_(20)=225` …(Given)
`t_(n)=a+(n-1)d` …(Formula)
`t_(18)=a+17d,t_(19)=a+18d and t_(20)=a+19d.`
`therefore(a+17d)+(a+18d)+(a+19d)=225`.
`therefore3a+54d=225`
`thereforea+18d=75` ...(Dividing both sides by 3)...(1) The sum of the last three terms, i.e.
`t_(35)+t_(36)+t_(37)=429` ...(Given)
`t_(35)=a+34d,t_(36)=a+35d,t_(37)=a+36d`
`therefore(a+34d)+(a+35d)+(a+36d)=429`
`therefore a+35d=143` ...(2)
Substracting equation (1) from equation (2),
`a+35d=143` ...(2)
`a+18d=75` ...(1)
`(---)/(17d=68)`
`therefored=(68)/(17)" "therefored=4`
Substituting d = 4 in equation (1),
`a+18(4)=75" "thereforea+72=75`
`thereforea=75-72 " "thereforea=3`
`t_(1)=a=3,t_(2)=t_(1)+d=3+4=7`
`t_(3)=t_(2)+d=7+4=11`
37th term `=a+36d`
`=3+36xx4`
`=3+144=147`