The sum of first n terms of an arithmetic progression is 210 and sum o

1.	The sum of first n terms of an arithmetic progression is 210 and sum of its first (n-1) terms is 171.If the first term is 3, then write the arithmetic progression.
Answer» Step-by-step explanation: Let a be the 1st term , d be the common difference and Sn be the sum of the 1st N terms of an A.P. ,acordingly:- Sn=210. , S(n-1)=171. , a=3. , d=? Sn = n/2.[2a+(n-1).d]=210. ………………(1) But. tn= Sn-S(n-1). or. a+(n-1).d = 210–171. or. 3+(n-1).d= 39 or. (n-1). d=36. putting (n-1).d=36 in EQN. (1) n/2.[2×3 +36]=210 or. n/2.[6+36]=210 => n = (210×2)/42 =10. but d = 36/(n-1) = 36/(10–1) = 4. a=3 and d =4 . then A.P. is:- 3 , 7 , 11 , 15 , 19 , ………….. Answer. helpful to you

The sum of first n terms of an arithmetic progression is 210 and sum of its first (n-1) terms is 171.If the first term is 3, then write the arithmetic progression.

Answer»

Step-by-step explanation:

Let a be the 1st term , d be the common difference and Sn be the sum of the 1st

N terms of an A.P. ,acordingly:-

Sn=210. , S(n-1)=171. , a=3. , d=?

Sn = n/2.[2a+(n-1).d]=210. ………………(1)

But. tn= Sn-S(n-1).

or. a+(n-1).d = 210–171.

or. 3+(n-1).d= 39

or. (n-1). d=36. putting (n-1).d=36 in EQN. (1)

n/2.[2×3 +36]=210

or. n/2.[6+36]=210 => n = (210×2)/42 =10.

but d = 36/(n-1) = 36/(10–1) = 4.

a=3 and d =4 . then A.P. is:-

3 , 7 , 11 , 15 , 19 , ………….. Answer.