Advertisements
Advertisements
Question
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is
Options
13
9
21
17
Advertisements
Solution
In the given problem, the sum of three consecutive terms of an A.P is 51 and the product of the first and the third terms is 273.
We need to find the third term.
Here,
Let the three terms be (a-d),a,(a + d) where, a is the first term and d is the common difference of the A.P
So,
(a - d) + a + ( a + d) = 51
3a = 51
`a = 51/3`
a = 17
Also,
( a - d) ( a + d ) = 273
a2 - d2 = 273 [ Using a2 - d2 = (a + b ) (a - b)]
172 - d2 = 273
289 - d2 = 273
Further solving for d,
Now, it is given that this is an increasing A.P. so d cannot be negative.
So, d = 4
Substituting the values of a and d in the expression for the third term, we get,
Third term = a + d
So,
a + d = 17 + 4
= 21
Therefore, the third term is a3 = 21
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.
Find the sum of the following APs:
2, 7, 12, ..., to 10 terms.
In an AP given an = 4, d = 2, Sn = −14, find n and a.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
Find the sum of the following arithmetic progressions:
`(x - y)/(x + y),(3x - 2y)/(x + y), (5x - 3y)/(x + y)`, .....to n terms
Find the sum of all integers between 84 and 719, which are multiples of 5.
Find the three numbers in AP whose sum is 15 and product is 80.
Divide 24 in three parts such that they are in AP and their product is 440.
If 18, a, (b - 3) are in AP, then find the value of (2a – b)
If (2p – 1), 7, 3p are in AP, find the value of p.
In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.
If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164?
Sum of n terms of the series `sqrt2+sqrt8+sqrt18+sqrt32+....` is ______.
The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is
The common difference of the A.P. is \[\frac{1}{2q}, \frac{1 - 2q}{2q}, \frac{1 - 4q}{2q}, . . .\] is
Find the sum of first 1000 positive integers.
Activity :- Let 1 + 2 + 3 + ........ + 1000
Using formula for the sum of first n terms of an A.P.,
Sn = `square`
S1000 = `square/2 (1 + 1000)`
= 500 × 1001
= `square`
Therefore, Sum of the first 1000 positive integer is `square`
Find S10 if a = 6 and d = 3
How many terms of the AP: –15, –13, –11,... are needed to make the sum –55? Explain the reason for double answer.
Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
