Advertisements
Advertisements
Question
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
Advertisements
Solution
The numbers from 50 to 350 which are divisible by 6 are 54, 60, 66, ……, 348.
∴ First term =a=t1= 54, d= 6 and tn= 348
tn = a+(n-1)d
∴348 = 54+(n-1)6
∴294 = (n-1)6
∴49 = n-1
∴n = 50
`S_n= n/2(t_1+t_n)`
∴S50= `50/2(54+348)`
=25*402
=10050
t15 = 54+14(6)= 54+84 = 138
Thus, the sum of all numbers from 50 to 350,which are divisible by6, is 10050 and the 15th term of this A.P. is 138.
APPEARS IN
RELATED QUESTIONS
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take `pi = 22/7`)
[Hint: Length of successive semicircles is l1, l2, l3, l4, ... with centres at A, B, A, B, ... respectively.]
Find the sum of all even integers between 101 and 999.
Find the sum of the first 40 positive integers divisible by 5.
Determine k so that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
Write the next term of the AP `sqrt(2), sqrt(8), sqrt(18),`....
The sum of the first n terms of an AP is (3n2 + 6n). Find the nth term and the 15th term of this AP.
How many terms of the A.P. 21, 18, 15, ... must be added to get the sum 0?
Find the sum of the first 15 terms of each of the following sequences having nth term as xn = 6 − n .
Find the sum of all 2-digit natural numbers divisible by 4.
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then \[\frac{S_1}{S_2} =\]
Q.5
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.
The sum of first 16 terms of the AP: 10, 6, 2, ... is ______.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a + c)(b + c - 2a))/(2(b - a))`
Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?
