Advertisements
Advertisements
Question
Find the sum of first 8 multiples of 3
Advertisements
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
Find the sum of the following APs.
0.6, 1.7, 2.8, …….., to 100 terms.
Find the sum of the following arithmetic progressions:
−26, −24, −22, …. to 36 terms
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
If \[\frac{5 + 9 + 13 + . . . \text{ to n terms} }{7 + 9 + 11 + . . . \text{ to (n + 1) terms}} = \frac{17}{16},\] then n =
Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th installments of loan? What amount of loan she still has to pay after the 30th installment?
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
