Find the sum of $20$ terms of an arithmetic progression, if its first term is $2$ and the seventh term is $20$ .

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Solution

First term $(a) = 2$
Seventh term $(a + (7 - 1) d) = 20$ [As the terms in $A P$ ]
So, $2 + 6 d = 20 6 d = 18, d = 3$
Sum of $20$ terms
$= \frac{n}{2} (2 a + (n - 1) d)$ [where $n = 20, a = 2$ and $d = 3$ ]
$= \frac{20}{2} (2 (2) + (20 - 1) 3) = 10 (4 + (19 \times 3)) = 10 + (4 + 57) = 610$
Sum of first $20$ terms $= 610$

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Find the sum of 20 terms of an arithmetic progression, if its first term is 2 and the seventh term is 20.

First term(a)=2Seventh term (a+(7−1)d)=20 [As the terms in AP]So,2+6d=206d=18,d=3Sum of 20 terms=n2(2a+(n−1)d) [where n=20,a=2 and d=3]=202(2(2)+(20−1)3)=10(4+(19×3))=10+(4+57)=610Sum of first 20 terms=610

Find the sum of $20$ terms of an arithmetic progression, if its first term is $2$ and the seventh term is $20$ .