Find the sum:$3 + 11 + 19 + ………. + 803$

Given:

Given sequence is $3 + 11 + 19 + ………. + 803$.

To do:

We have to find the sum of $3 + 11 + 19 + ………. + 803$.

Solution:

Here,

\( 3+11+19+\ldots+803 \) is in A.P.
\( a=3, d=11-3=8 \) and \( l=803 \)

We know that,
\( a_{n}=a+(n-1) d \)

\( \Rightarrow 803=3+(n-1) \times 8 \)
\( \Rightarrow 803=3+8 n-8 \)
\( \Rightarrow 8 n=803+8-3=808 \)
\( n=\frac{808}{8}=101 \)

\( \mathrm{S}_{n}=\frac{n}{2}[a+l] \)
\( =\frac{101}{2}[3+803] \)

\( =\frac{101}{2} \times 806 \)
\( =101 \times 403=40703 \)

Therefore, the sum of the given sequence is 40703. 

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Updated on: 10-Oct-2022

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