Advertisements
Advertisements
Question
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
Options
501th
502th
508th
none of these
Advertisements
Solution
In the given problem, let us take the first term as a and the common difference as d.
Here, we are given that,
`a_9 = 449` ...............(1)
`a_449 = 9 ` .................(2)
We need to find n
Also, we know,
`a_n = a + ( n- 1) d`
For the 9th term (n = 9),
`a_9 = a + ( 9 -1) d`
449 = a + 8d (Using 1 )
a = 449 - 8 d .................(3)
Similarly, for the 449th term (n = 449),
`a_449 = a + ( 449 - 1 )d`
9 = a + 448d (Using 2 )
a = 9 - 448 d .............(4)
Subtracting (3) from (4), we get,
a -a = ( 9 - 448d) - ( 449 - 8d)
0 = 9 - 448d - 449 + 8d
0 = -440 - 440d
440d = - 440
d = - 1
Now, to find a, we substitute the value of d in (3),
a = 449 - 8 (-1)
a = 449 + 8
a = 457
So, for the given A.P d = - 1 and a = 457
So, let us take the term equal to zero as the nth term. So,
`a_n = 457 + ( n- 1) ( -1 ) `
0 = 457 - n + 1
n = 458
So, n = 458
APPEARS IN
RELATED QUESTIONS
In an A.P., if S5 + S7 = 167 and S10=235, then find the A.P., where Sn denotes the sum of its first n terms.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]
A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 `1/2` m apart, what is the length of the wood required for the rungs?
[Hint: number of rungs = `250/25+ 1`]
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Choose the correct alternative answer for the following question .
In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Sum of n terms of the series `sqrt2+sqrt8+sqrt18+sqrt32+....` is ______.
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
The term A.P is 8, 10, 12, 14,...., 126 . find A.P.
Q.7
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
Solve for x: 1 + 4 + 7 + 10 + ... + x = 287.
What is the sum of an odd numbers between 1 to 50?
A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
The sum of first five multiples of 3 is ______.
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
Find the sum of the integers between 100 and 200 that are not divisible by 9.
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
The sum of 41 terms of an A.P. with middle term 40 is ______.
