Find the sum of all two digits odd multiples
of 3
Answers
Answer:
First two digit number divisible by 3 = 12
Last two digit number divisible by 3 = 99
∴ First term, (a)=12
Common difference, (d)=3
Last term, (l)=99
n=?
As we know that,
a
n
=a+(n−1)d
∴99=12+(n−1)3
⇒(n−1)
3
87
⇒n=29+1=30
∴ Sum of n terms of an A.P., when last term is known is given by-
S
n
=
2
n
(a+l)
∴S
30
=
2
30
(12+99)
⇒S
30
=15×111=1665
Step-by-step explanation:
answer is (1665)
Two digit odd multiples of 3 numbers are =15,21,27,33,39,45,51,57,63,69,75,81,87,93,99
therefore,
the sum of the number
S_{n}=\frac{n}{2}(2\times n+(n-1)\times d)S
n
=
2
n
(2×n+(n−1)×d)
Where, n=No. of terms=15
d= difference between two consecutive number=6
\begin{gathered}\therefore S_{n}=\frac{n}{2}[2\times n+(n-1)\times d]\\\\\Rightarrow S_{n}=\frac{15}{2}[2\times 15+(15-1)\times 6]\\\\\Rightarrow S_{n}=\frac{15}{2}[30+(14\times 6)]\\\\\Rightarrow S_{n}=\frac{15}{2}[30+84]\\\\\Rightarrow S_{n}=\frac{15}{2}\times 114\end{gathered}
∴S
n
=
2
n
[2×n+(n−1)×d]
⇒S
n
=
2
15
[2×15+(15−1)×6]
⇒S
n
=
2
15
[30+(14×6)]
⇒S
n
=
2
15
[30+84]
⇒S
n
=
2
15
×114
\begin{gathered}\\\\\Rightarrow S_{n}=855\end{gathered}
⇒S
n
=855