Question

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

Answer 1

Answer:

here, a=3

d=n-(n-1)=>210-171=39

Step-by-step explanation:

3,42,81,120,159,.......

mark me brainliest!!!....hope its helpful!!

Answer 2

$\bf{ \underline{ \underline{ \: Answer}} }: - \\ \\ \underline{\underline{\bf{{Step - by - step \: explanation}}}} : - \\ \\ \underline{\bf{given}} : -$

Sum of first n th term of Airthmatic progression is 210.

Sum of first (n -1) th term is 171 .

First term of this AP (a) is 3

We know that , if AP has n terms then after removing (n - 1) terms ,left last term.

Last term (l) = 210 - 171 = 39,

first term (a) = 3

$\bf{sum \: of \: n \: th \: term \: = \frac{n}{2} \big(a \: + l \big)} \\ \\ \bf{\implies \: 210 = \frac{n}{2} \big(3 + 3 \big)} \\ \\ \implies \: \bf{420 = 42 \: n} \\ \\ \implies \: \bf{n = 10}$

Now , According to the formula of nth term of AP is -

$\bf{sn = \frac{n}{2} \bigg(2a + (n - 1)d \bigg) }\\ \\ \bf{ \implies \: 210 = \frac{10}{2} \bigg(2 \times 3 + 9d \bigg)} \\ \\ \implies \: \bf{\frac{210}{5} = 6 + 9d} \\ \\ \implies \: \bf{36 = 9d} \\ \\ \implies \: \bf{d = 4}$

Hence, Common difference (d) = 4

Therefore,

Required AP is → 3 ,3+4 ,3+2×4 ,.....,39

→ 3, 7 ,11, 15, .....,39

Hope it helps you.