Question

An Arithmetic progression consists of 37 terms.The sum of first 3 terms is 12 and the sum of last 3 terms is 318,then find the first and last term of the AP​ ​

Answer 1

Answer:

n=37

Sum of 3 terms =12

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

S3=3/2(2a+2d)

12=3/2×2(a+d)

12/3=a+d

4=a+d (make it first equation)

and by following these same steps make second equation with the information given in the question

Answer 2

Answer:

Let a,d and n be the first term, common difference and the no. of terms of given AP.

Here n=37

so middle most term is

2

n+1

⇒

2

37+1

⇒18

th

∴ three middle most terms are 18

th

,19

th

,20

th

a

18

+a

19

+a

20

=225⇒a+17d+a+18d+a+19d=225

⇒3a+54d=225 ...(1)

last three terms are 35

th

,36

th

,37

th

a+34d+a+35d+a+36d=429

⇒3a+105d=429 ...(2)

subtracting (1) from (2), we get

51d=204

⇒d=4

putting d in (1), we get

3a+54×4=225

3a=225−216⇒a=3

Therefore the AP is a,a+d,a+2d,a+3d,....

i.e 3,7,11,15...

Explanation:

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YOU ARE IN CLASS 10..

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