Question

Sum of the first n terms of an arithmetic progression is 210and sum of the first n-1 terms of an arithmetic progression is 171.if the first term is 3 then write the arithmetic progression

Answer 1

Answer:

Arithmetic progression is 3,7,11,15... 39

Step-by-step explanation:

To find the AP if the sum of first n terms of arithmetic progression is 210 and sum of its first (n-1) terms is 171. If the first term is 3 .

As we know that if AP has n terms,then after removing (n-1) terms,we are left with the last term.

Difference of sum of n terms-difference of sum of (n-1) terms= last term

l= 210-171

l= 39

a= 3

$S_n = \frac{n}{2} (a + l) \\ \\ 210 = \frac{n}{2} (3 + 39) \\ \\ 420 = 42n \\ \\ n = \frac{420}{42} \\ \\ n = 10$

now as we know the AP has n terms so put the formula of n terms as shown under to find the value of d

$S_n = \frac{n}{2} (2a + (n - 1)d) \\ \\ 210 = \frac{10}{2} (2 \times 3 + (10 - 1)d) \\ \\ 210 = 5(6 + 9d) \\ \\ \frac{210}{5} = (6 + 9d) \\ \\ 42 - 6 = 9d \\ \\ 36 = 9d \\ \\ d = 4$

common difference d= 4

so, arithmetic progression is 3,7,11,15... 39

Hope it helps you.

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.