Question

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 3, them write the arithmetic progression

Answer 1

Answer:

3 , 7 , 11 , 15 , 19 , 23

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Step-by-step explanation:

We know that nth term tn = sum of of the first n terms - sum of the first n-1 terms= 210–171 = 39.

But tn = a+ (n-1)d = 39 => 3+(n-1)d = 39

=> (n-1)d = 39–3= 36. ………(1)

Now And = (n/2){2a+(n-1)d} = 210

=> (n/2){6+36} = 210 (using (1))

=> (n/2)(42)= 210 => n/2 = 210/42=5 => n= 2×5=10

Using n=10 in(1) we get (10–1)d = 36

=> d = 36/9= 4

3 , 7 , 11 , 15 , 19 ,23 ………….. Answer.