Question

An arithmetic progression has 41 terms. The sum of three middle terms is 237 and the sum of the last terms is 465 write the progression ​

Answer 1

The arithmetic progression is : -1,3,7,11,15,19,23,27,31,35,39,43,47,51,55,59..........153,157,161

•Let , the first term be a,

•No. of terms = n

•common difference = d

•We have, nth term, a(n) = a + (n-1)d

•Sum upto n terms,

Sn = n/2 {2a + (n-1)d}

•Since n= 41, the middle terms are a20, a21 and a22.

•And the last 3 terms are a39, a40, a41

•S(middle) = 3/2 {2*a20 +(3-1)d}

•S(last) = 3/2 {2*a39 +(3-1)d}

•According to the question,

•237 = 3/2 {2*a20 +2d}

and 465 = 3/2 {2*a39 +2d}

•Substracting them, we get

465-237 = 3/2 {2*a39 +2d- 2*a20 - 2d}

• 228 = 3/2(2*a39 - 2*a20)

• 228/3 = a39- a20

• 76 = a39- a20

• 76 = a + 38d - (a + 19d)

• 76 = 19 d

• d = 76/19 = 4

•Putting this value in sum of middle terms, we get

237 = 3/2 (2*a20 + 2*4)

• 237/3 = a20 + 4

• a20 = 79 -4 = 75

• a + 19*4 = 75

• a = 75 - 76

• a = -1.

Done