In an ap the sum of n terms is 1/5(5n^2+3n).find its 40th term
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Sn=1/5(5n^2+3n)
here put n =40 and n=39
tn =Sn-S(n-1)
Here,
t40=S40-S39
=1/5[5(40)^2+3(40)]-1/5[5(39)^2+3(39)]
=1/5[5(1600)+120]-1/5[5(1521)+117]
=1/5(8000+120)-1/5(7605+117)
=1/5(8120)-1/5(7722)
=(8120-7722)/5
=398/5=79.6
here put n =40 and n=39
tn =Sn-S(n-1)
Here,
t40=S40-S39
=1/5[5(40)^2+3(40)]-1/5[5(39)^2+3(39)]
=1/5[5(1600)+120]-1/5[5(1521)+117]
=1/5(8000+120)-1/5(7605+117)
=1/5(8120)-1/5(7722)
=(8120-7722)/5
=398/5=79.6
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