Find the sum of the terms of A.P 2,6,10,14....,32.
Answers
Answered by
12
a = 2, d = 6 - 2 = 4, n = 32
Sum of n terms = sn = n/2[2a+(n-1)d]
Substituting = 32/2[2(2)+(32-1)4]
= 16[4+(31)4]
= 16[4+124]
= 16 * 128
= 2048
Hence Sum of terms of AP 2,6,10,14..........,32. is 2048
Answered by
8
hey the question is wrong as 32 can't be the term as we add 4 consecutively we will get 18,22,26,30,34 see there's no 32
and if you meant that 32 is the number of term than
s=n/2(2a+(n-1)*d)
32/2(4+31*4)
16*(128)
2^4*2^7*(for easier calculation)
=2^11
2048
(*this is answer is only valid when number of term is 32 else the question is wrong)
and if you meant that 32 is the number of term than
s=n/2(2a+(n-1)*d)
32/2(4+31*4)
16*(128)
2^4*2^7*(for easier calculation)
=2^11
2048
(*this is answer is only valid when number of term is 32 else the question is wrong)
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