Advertisements
Advertisements
प्रश्न
How many terms of the A.P. 27, 24, 21, .... should be taken so that their sum is zero?
Advertisements
उत्तर
The given AP is 27, 24, 21, ..
First term of the AP = 27
Common difference = 24 − 27 = −3
Let the sum of the first x terms of the AP be 0.
Sum of first x terms = `x/2`[2×27+(x−1)(−3)]=0
⇒`x/2`[54+(−3x+3)]=0
⇒x(54−3x+3)=0
⇒x(57−3x)=0
Now, either x = 0 or 57 − 3x = 0.
Since the number of terms cannot be 0, x≠0.
∴ 57 − 3x = 0
⇒ 57 = 3x
⇒ x = 19
Thus, the sum of the first 19 terms of the AP is 0.
APPEARS IN
संबंधित प्रश्न
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its first 20 terms.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].
The first term of an A.P. is p and its common difference is q. Find its 10th term.
If `4/5` , a, 2 are three consecutive terms of an A.P., then find the value of a.
In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)
The sum of first n terms of the series a, 3a, 5a, …….. is ______.
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
