Advertisements
Advertisements
Question
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
Advertisements
Solution
For an A.P.
t4 = 11
`\implies` a + 3d = 11 ...(i)
Also, t8 – 2t4 = 5
`\implies` (a + 7d) – 2 × 11 = 5
`\implies` a + 7d – 22 = 5
`\implies` a + 7d = 27 ...(ii)
Subtracting equation (i) from equation (ii), we get
a + 3d = 11
a + 7d = 27
– – –
– 4d = – 16
d = `(- 16)/(- 4)`
`\implies` d = 4
Substituting d = 4 in equation (i), we get
a + 3d = 11
a + 3 × 4 = 11
`\implies` a + 12 = 11
`\implies` a = –1
∴ Required A.P. = a, a + d, a + 2d, a + 3d, ....
∴ a = –1
∴ a + d = –1 + 4 = 3
∴ a + 2d = –1 + 2(4) = –1 + 8 = 7
∴ a + 3d = –1 + 3(4) = –1 + 12 = 11
Sum of first 50 terms of this A.P.
=`50/2 [2 xx (-1) + 49 xx 4]`
= 25[–2 + 196]
= 25 × 194
= 4850
RELATED QUESTIONS
How many terms of the A.P. 27, 24, 21, .... should be taken so that their sum is zero?
Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
What value is generated in the above situation?
First term and the common differences of an A.P. are 6 and 3 respectively; find S27.
Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
If the seventh term of an A.P. is \[\frac{1}{9}\] and its ninth term is \[\frac{1}{7}\] , find its (63)rd term.
If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
x is nth term of the given A.P. an = x find x .
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
The sum of the first 2n terms of the AP: 2, 5, 8, …. is equal to sum of the first n terms of the AP: 57, 59, 61, … then n is equal to ______.
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]
In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.
