Question

Find the sum of all two digit odd multiples of 3

Answer 1

HEY MATE YOUR ANSWER IS HERE....

The two digit odd multiples of 3 are....

15, 21, 27, 33.........99

Here, a(first term) =15;
Common difference(d) =6,
tn=99

Now,

$tn = a + (n - 1)d \\ 99 = 15 + (n - 1) \times 6 \\ 99 = 15 + 6n - 6 \\ 99 = 9 + 6n \\ 99 - 9 \div 6 = n \\ 90 \div 6 = n \\ hence \: n = 15$

Hence there are 15 odd multiples divisible by 3.

TO FIND : Sum of all these two digit multiples.

By formula,

$sn \: = \: n \div 2(a \: + tn) \\ sn = 15 \div 2(15 + 99) \\ sn = 15 \div 2 \times 114 \\ sn = 15 \times 57 \\ therefore.sn = 855$

Hence the sum of two digit odd multiples of 3=

$855$

$<marquee>HOPE IT HELPS YOU OUT$

Answer 2

Answer:

The required sum = 855

Step-by-step explanation:

Formula used:

The number of terms of an A.P

a, a+d, a+2d...... is

$n=\frac{l-a}{d}+1$

The sum of n terms of an A.P

a, a+d, a+2d..............is

$S_n=\frac{n}{2}[a+l]$

Two digit odd multiples of 3 are

15, 21, 27, 33....................99

This is an A.P with a=15, d=6

$n=\frac{l-a}{d}+1\\\\n=\frac{99-15}{6}+1\\\\n=\frac{84}{6}+1\\\\n=14+1\\\\n=15$

Therefore,

The required sum

$=S_15\\\\=\frac{n}{2}[a+l]\\\\=\frac{15}{2}[15+99]\\\\=\frac{15}{2}[114]\\\\=(15)[57]\\\\=855$