the sum of first five multiples of 3 in A.P
Answers
Answered by
45
First five multiple of 3 are:
3, 6, 9, 12, 15
Here first term, a = 3
common difference, d = 6 – 3 = 3
Number of term, n = 5
Sn = n/2 [2a+ (n-1)d]
s5 =5/2 [2×3 + (5-1) 3]
=5/2 [6+4× 3]
= 5/2× 18
⇒ s5 = 9 × 5 = 45
Hope this helps you ¡!!!!!!!!!!!!!!¡
3, 6, 9, 12, 15
Here first term, a = 3
common difference, d = 6 – 3 = 3
Number of term, n = 5
Sn = n/2 [2a+ (n-1)d]
s5 =5/2 [2×3 + (5-1) 3]
=5/2 [6+4× 3]
= 5/2× 18
⇒ s5 = 9 × 5 = 45
Hope this helps you ¡!!!!!!!!!!!!!!¡
Answered by
10
Answer:
Step-by-step explanation:
First 5 multiples of 3 are as follows:
3,6,9,12,15
Now, we got a=3,d=3 and no. Of terms =5
Putting the values in
Sn=n/2[2a+(n-1)d]
S5 =5/2 [2×3 + (5-1) 3]
=5/2 [6+4× 3]
= 5/2× 18
⇒ S5 = 9 × 5 = 45
Hence 45 is the answer to the question
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