Question

an AP consist of 37 terms the sum of the first 3 terms of it is 12 and the sum of last 3 terms is 318 find the first and last terms​

Answer 1

The AP has 37 terms.

The middle term is the 19th term.

So the sum of the middle three terms = (a+17d)+(a+18d)+(a+19) = 225, or

3a + 54d = 225, or

a+18d = 75 …(1)

So the sum of the last three terms = (a+34d)+(a+35d)+(a+36) = 429, or

3a+105 = 429, or

a + 35d = 143 …(2)

Subtract (1) from (2)

17d = 143–75 = 68, or d = 68/17 = 4.

From (1), a = 75–18d = 75–72 = 3.

So the first term of the AP is 3 and the cd = 4. The AP is thus: 3,7,11,15,…

Let the middle term of the AP be a and common difference is d.

The Series is : a−18d,a−17d,1–16d,...,a−d,a,a+d,a+2d,…,a+17d,a+18d

Total number of terms = 37

Sum of the 3 middle terms = a−d+a+a+d=225⟹3a=225⟹a=75

Sum of last 3 terms = 429⟹a+16d+a+17d+a+18d=429

⟹3a+51d=429⟹51d=42−3a=429−225=204

⟹d=4

∴ the Series:

75−(18∗4),75−(17∗4),75−(16∗4),…,75+(18∗4)

⟹3,7,11,…,147

Since there are 37 terms. So, middle most term = (37 + 1)/2 = 19th term

=> 3 middle most terms are

18th, 19th & 20th

a = first term : d = common difference

T18 = a + 17d

T19 = a + 18d

T20 = a + 19d

=> 3a + 54 d = 225 …….. (1)

Now, last 3 terms..

T35 = a+ 34d

T36 = a + 35d

T37 = a + 36d

=> 3a + 105d = 429 ………….. (2)

Eq (1) - eq(2)

51d = 204

=> d = 4

Hence, 3a + 216 = 225

=> a = 3

So, AP is…

3,7,11,15,19,…………..147