Question

if the value of first term in an AP is 2 and the nth term is 41, and Sn is 860 then find the value of n (number of terms in the AP) ​

Answer 1

a=2
Tn=41
Sn=860

Tn=a+(n-1)d
41=2+(n-1)d
41-2=(n-1)d
(n-1)d=39

Sn=n/2(2a+(n-1)d)
860=n/2(2×2+39). [put the value (n-1)d=39]
860×2=n(43)
43n=1720
n=1720/43
n=40

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Answer 2

Given:

First term,a=2

nth term= 41

Sn=860

To Find:

The value of n?

Step-by-step explanation:

• It is provided nth term of A.P is 41

• This imply

               $\bf a_n=a+(n-1)d\\\\41=2+(n-1)d\\\\(n-1)d=39$

Now we have Sn=860

$\bf\Rightarrow Sn=\frac{n}{2}(2a+(n-1)d)\\\\\Rightarrow860=\frac{n}{2}(2\times2+39)\\\Rightarrow860\times2=43n\\\Rightarrow\frac{1720}{43}=n\\\Rightarrow n=40$

Hence, value of n is 40