Question

OR
The first term of an A.P. is 5, the last term is 45 and sum is 400. Find the number of em
common difference.​

Answer 1

Answer:

8/3

Step-by-step explanation:

Answer 2

❏ Formula Used:-

If in an A.P. series "a" be the first term and "d" be the common difference then ,

(1) The n'th term is given by the formula .

$\sf\longrightarrow\boxed{ a_n=a+(n-1)d }$

(2)Sum of n number of terms ,

$\sf\longrightarrow\boxed{ S_n=\frac{n[2a+(n-1)d]}{2} }$

❏ Question:-

Q) The first term of an A.P. is 5, the last term is 45 and sum is 400. Find the number of em

The first term of an A.P. is 5, the last term is 45 and sum is 400. Find the number of emcommon difference.

❏ Solution:-

❚ Given➝

•First term(a)=05

•Last term(l) =45

•Sum($S_n$)=400

❚ To Find➝

•Number of terms(n)=?

•Common difference (d)=?

Now, using the above last term formulas,

$\sf\longrightarrow l=a+(n-1)d$

$\sf\longrightarrow 45=5+(n-1)d$

$\sf\longrightarrow 45-5=(n-1)d$

$\sf\longrightarrow\boxed{ 40=(n-1)d}\:\:...........(i)$

$\sf\longrightarrow \frac{40}{d}=(n-1)$

$\sf\longrightarrow \frac{40}{d}+1=n$

$\sf\longrightarrow \boxed{n=\frac{40}{d}+1}$

Now, Again using the Sum Formula,

$\sf\longrightarrow S_n=\frac{n[2a+(n-1)d]}{2}$

$\sf\longrightarrow 400=\frac{n[2\times5+(n-1)d]}{2}$

$\sf\longrightarrow 400\times2=n[10+40]$

[ using equation (i) ]

$\sf\longrightarrow 800=50n$

$\sf\longrightarrow n =\frac{\cancel{800}}{\cancel{50}}$

$\sf\longrightarrow\boxed{\large{\red{ n=16}}}$

Now,

$\sf\longrightarrow 40=(n-1)d$

$\sf\longrightarrow 40=(16-1)d$

$\sf\longrightarrow 40=15d$

$\sf\longrightarrow d=\frac{40}{15}$

$\sf\longrightarrow \boxed{\large{\red{d=\frac{8}{3}}}}$

∴ Number of terms(n) = 16

∴ Common difference(d)=$\frac{8}{3}$

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