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 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

last term(l) = 210-171

l= 39

a= 3

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

S

n

=

2

n

(a+l)

210=

2

n

(3+39)

420=42n

n=

42

420

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

\begin{lgathered}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 \\ \\\end{lgathered}

S

n

=

2

n

(2a+(n−1)d)

210=

2

10

(2×3+(10−1)d)

210=5(6+9d)

5

210

=(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.