Question

If the sum of n terms of AP is 210 and sum of n-1 term of AP is 171. If 1st term of an ap is 3, find AP ​

Answer 1

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

• 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.

Answer 2

Answer:

• Sum of n terms = 210
• Sum of n-1 terms = 171
• First term = 3

Now, sum of n terms - sum of n -1 terms = last term.

So, l = 210 - 171

➸ last term = l = 39

Also, given, a = 3.

The formula to find the sum of n terms when the last term is given = n/2(a + l)

Hence, 210 = n/2(3 + 39)

➸ 210 = n/2(42)

➸ 210 = n * 21

➸ n = 210/21

➸ n = 10

Now, with the other formula to find sum of n terms, we can find common difference.

The Formula is : Sn = n/2 (2a + (n-1)d)

So, 210 = 10/2 (2*3 + (10-1)d)

➸ 210 = 5 (6 + 9d)

➸ 210 = 30 + 45d

➸ 210 = 30 + 45d

➸ 210 - 30 = 45d

➸ 180 = 45d

➸ d = 180/45

➸ d = 4

Hence, the common difference = 4.

So, arithmetic progression = 3, 3+4, 3+8, 3+12, .....39

AP = 3, 7, 11, 15,.... 39