If tn=3+4n then find the AP and hence find the sum of its first 15 terms
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Answered by
0
Answer:
t1=3+4.1=7
t2=3+4.2=11
t3=3+4.3=15
now
s15=15/2(7+(15-1)4)
15/2(7+56)
15/2(63)
Answered by
0
Answer:
525
Step-by-step explanation:
tₙ = 3 + 4n
tₙ = t + (n - 1)d
∴ t₁ = 3 + 4 x 1 = 7
t₁₅ = 3 + 4 x 15 = 63
so , sum of first 15 terms
=> Sₙ = n/2 ( t₁ + tₙ )
S₁₅ = 15/2 (7 + 63) = 15/2 x 70
= 15 x 35 [the 70 got dvided by 2]
= 525
Hope it helped ! please mark as brainliest
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