Question

the sum of first n terms of an ap is 210 and the sum of its first N -1 term is 175 if the first term 3 the write the AP​

Answer 1

Correction required:

Sum of the first (n - 1) terms is 171

Solution:

Let a and d be the first term and the common difference respectively.

Given, Sₙ = 210 and Sₙ_₁ = 171

Then

n/2 * {2a + (n - 1)d} = 210 ... (1)

& (n - 1)/2 * {2a + (n - 2)d} = 171 ... (2)

So nth term = Sₙ - Sₙ_₁ = 210 - 171

or, a + (n - 1)d = 39

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

since a = 3 (given)

or, (n - 1)d = 39 - 3

or, (n - 1)d = 36 ... (3)

Using (3), from (1), we get

n/2 * (6 + 36) = 210, since a = 3

or, n/2 * 42 = 210

or, n = 210 * 2 / 42

or, n = 10

Thus the given AP contains 10 terms.

From (3), we get

(10 - 1) d = 36

or, 9d = 36

or, d = 36/9

or, d = 4

Thus the common difference is 4.

Therefore, the AP is

3, 3 + 4, 3 + 2 (4), 3 + 3 (4), ..., 39

i.e., 3, 7, 11, 15, ..., 39

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.