Q. 14.2( 217 Votes )
Find the sum of the following APs:
(i) 2, 7, 12, . . ., to 10 terms.
(ii) –37, –33, –29, . . ., to 12 terms.
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.
(iv) 
Answer :
(i) Here, a = 2, d = 5 and n = 10
Sum of n terms can be given as follows:
S10
= 5(4 + 45)
= 5
= 245
Thus, sum of the 10 terms of given AP (Sn)=245
(ii) Here, a = - 37, d = 4 and n = 12
Sum of n terms can be given as follows:
S12
= 6(-74 + 44)
= 6
=-180
Thus, sum of the 12 terms of given AP (Sn)= -180
(iii) Here, a = 0.6, d = 1.1 and n = 100
Sum of n terms can be given as follows:
S100
= 50(1.2 + 108.9)
= 50
=5505
Thus, sum of the 100 terms of given AP (S100) = 5505
(iv) Here,
a= , n = 11,
d =
Sum of n terms can be given as follows:
S11
= (
+
)
=
=
Thus, sum of the 100 terms of given AP (Sn)=
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