Advertisements
Advertisements
Question
Find the sum of first 8 multiples of 3
Solution 1
First, 8 multiples of 3 are { 3, 6, 9...,24}
We can observe they are in AP with first term (a) = 3 and last term (l) = 24 and number of terms are 8.
`S_n = n/2 (a + l)`
`=> S_n = 8/2(3 + 24)`
`S_8 = 4 xx (3 + 24) = 108`
Hence, the sum of the first 8 multiples of 3 is 108
Solution 2
First 8 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24
The above sequence is an A.P
a= 3, d= 3 and last term l = 24
`S_n = n/2 (a + 1) = 8/2[3 + 24] = 4(27)`
`S_n = 108`
APPEARS IN
RELATED QUESTIONS
The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
In a flower bed, there are 43 rose plants in the first row, 41 in second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?
How many terms of the AP 21, 18, 15, … must be added to get the sum 0?
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
If 18, a, b, −3 are in A.P., the a + b =
x is nth term of the given A.P. an = x find x .
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
Find t21, if S41 = 4510 in an A.P.
