Question

35. An arithmetic progression consists of 37 terms. The sum of the first 3 terms of
it is 12 and the sum of its last 3 terms is 318, then find the first and last terms
of the progression.
OR​

Answer 1

Step-by-step explanation:

let the first three terms be

a-d, a, a+d

then last terms will be a-d+(37-1)d, a-d+(36-1)d, a-d+(35-1)d.

sum of first terms is given and is equal is 12

a-d+a+a+d=12

3a=12,

a=4 -----eq1

also sum of last three terms is 318

a+35d+a+34d+a+33d=318

3a+102d=310,

put the value of a from eq1

12+102d=310

102d=310

d=298/102.

d=2.921

first three terms are a-d, a, a+d = 1.078, 4, 6.921

last three terms are a+33d, a+34d, a+33d = 4+33*2.921 solve it calculator please

hope it helps

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