If 2-5-8-11-upto n terms- 610- then the value of n is

Given: 2+5+8+⋯upto n terms= 610 The given sequence is an A.P. with nbsp;a = 2, d nbsp; =3 Now, sum of n terms of A.P. is = n2[4+3(n 1)]=610 ⇒ 4n+3n(n 1) ...

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Standard XI

Mathematics

Sum of n Terms

If 2+5+8+11+⋯...

Question

If $2 + 5 + 8 + 11 + \dots upto n terms = 610$ , then the value of $n$ is

19

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Solution

The correct option is D $20$
Given: $2 + 5 + 8 + \dots upto n terms = 610$
The given sequence is an A.P. with $a = 2, d = 3$
Now, sum of $n$ terms of A.P. is
$= \frac{n}{2} [4 + 3 (n - 1)] = 610 \Rightarrow 4 n + 3 n (n - 1) = 1220 \Rightarrow 3 n^{2} + n - 1220 = 0 \Rightarrow 3 n^{2} + 61 n - 60 n - 1220 = 0 \Rightarrow (n - 20) (3 n + 61) = 0 ∴ n = 20 (∵ n > 0)$

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Sum of n Terms

Standard XI Mathematics

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If 2+5+8+11+⋯upto n terms=610, then the value of n is

The correct option is D 20Given: 2+5+8+⋯upto n terms=610 The given sequence is an A.P. with a=2, d=3 Now, sum of n terms of A.P. is =n2[4+3(n−1)]=610⇒4n+3n(n−1)=1220⇒3n2+n−1220=0⇒3n2+61n−60n−1220=0⇒(n−20)(3n+61)=0∴n=20 (∵n>0)

If $2 + 5 + 8 + 11 + \dots upto n terms = 610$ , then the value of $n$ is