find the sum of first 20terms of the arithmetic progression 3,8,13…. using the formula About the author Alice
Given:- An A.P = 3, 8, 13, . . . . To Find:- Sum of first 20 terms of an A.P. Formula Used:- [tex]{\boxed{\bf{S_n = \dfrac{n}{2}[2a+(n-1)d]}}}[/tex] Here, [tex]\bf S_n [/tex] = Sum of n terms a = First term of the A.P d = Common Difference n = No. of terms in an A.P Solution:- Using Formula, [tex]\bf :\implies\:S_n = \dfrac{n}{2}[2a+(n-1)d][/tex] Here, a = 3 d = 8 – 3 = 5 n = 20 Putting values, [tex]\sf :\implies\:S_{20} = \dfrac{20}{2}[2\times3+(20-1)5][/tex] [tex]\sf :\implies\:S_{20} = 10[6+19\times5][/tex] [tex]\sf :\implies\:S_{20} = 10[6+95][/tex] [tex]\sf :\implies\:S_{20} = 10\times101[/tex] [tex]\bf :\implies\:S_{20} = 1,010[/tex] Hence, The Sum of first 20 terms on given A.P is 1,010. ━━━━━━━━━━━━━━━━━━━━━━━━━ Reply
Given:-
To Find:-
Formula Used:-
Here,
Solution:-
Using Formula,
[tex]\bf :\implies\:S_n = \dfrac{n}{2}[2a+(n-1)d][/tex]
Here,
Putting values,
[tex]\sf :\implies\:S_{20} = \dfrac{20}{2}[2\times3+(20-1)5][/tex]
[tex]\sf :\implies\:S_{20} = 10[6+19\times5][/tex]
[tex]\sf :\implies\:S_{20} = 10[6+95][/tex]
[tex]\sf :\implies\:S_{20} = 10\times101[/tex]
[tex]\bf :\implies\:S_{20} = 1,010[/tex]
Hence, The Sum of first 20 terms on given A.P is 1,010.
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