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 is 3.then write the progression
Answers
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.
ACF, last term is 210-171
l= 39
a= 3
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
common difference d= 4
so, arithmetic progression is 3,7,11,15... 39
Hope it helps you.
Step-by-step explanation:
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.
ACF, last term is 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...
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