Please give the detailed solution of Q.31
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Answer:
Sn = (3n²/2) + 13n/2
S1 = 3/2 + 13/2 = 15/2
S1 = a = 15/2
Therefore the first term is 15/2
S2 = 3×4/2 + 13×2/2 = 6+13 = 19
a2 = S2 - S1 = 19 - 15/2 = 23/2
d = a2 - a1 = 23/2 - 15/2 = 8/2 = 4
Now we have, a = 15/2, d = 4, n = 25
a25 = a + 24d = 15/2 + 24×4 = 15/2 + 96
a25 = 207/2
#WMK
Answered by
1
Answer:
Answer:
Sn = (3n²/2) + 13n/2
S1 = 3/2 + 13/2 = 15/2
S1 = a = 15/2
Therefore the first term is 15/2
S2 = 3×4/2 + 13×2/2 = 6+13 = 19
a2 = S2 - S1 = 19 - 15/2 = 23/2
d = a2 - a1 = 23/2 - 15/2 = 8/2 = 4
Now we have, a = 15/2, d = 4, n = 25
a25 = a + 24d = 15/2 + 24×4 = 15/2 + 96
a25 = 207/2
#WMK
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