there are 27 terms in an A.P .sum of middle most term is 177 to the sum of last 3 terms is 321.find A.P
Answers
x1 = 7, x2 = 11, x3 = 15
Step-by-step explanation:
Let the 1st term = x1
for a.p., constant term added = n
∴ we can write equation
x2 = x1 + n
x3 = x1 + 2n
xn = x1 + (n-1)n .... (1)
There are total 27 terms
The middle term will be 14th term
∴ As per condition (1)
x13 + x14 + x15 = 177
substituting from equation 1 we get,
x13 = x1 + 12n
x14 = x1 + 13n
x15 = x1 + 14n
∴ x13 + x14 + x15 = (x1+12n) + (x1+13n) + (x1+14n) = 177 ..... (2)
∴3x1 + 39n = 177 .... (2)
For condition 2, sum of last three terms = 321
∴ x25 + x26 + x27 = 321
∴ (x1 + 24n) + (x1 + 25n) + (x1 + 26n) = 321
∴ 3x1 + 75n = 321 ...... (3)
∴ 3x1 = 321 - 75n..
substituting the value in equation 2, we get
(321 - 75n) + 39n = 177
∴ 321 - 177 = 75n - 39n
∴ 144 = 36n
∴ n = 4 .... (4)
Substtuting in eqn (3),
3x1 + 75× 4 = 321
∴ 3x1 + 300 = 321
∴ 3x1 = 21
∴ x1 = 7 .... (5)
1st 3 terms of a.p.
x1 = 7
x2 = x1 + n = 7 + 4 = 11
x3 = x2 + 2 = 11+ 4 = 15
∴ x1 = 7, x2 = 11, x3 = 15 .
Answer:
hey mate
your answer is in the below space
Step-by-step explanation:
Let the 1st term = x1
for a.p., constant term added = n
.. we can write equation
x2 = x1 + n
x3 = x1 + 2n
xn = x1 + (n-1)n
- (1)
There are total 27 terms
The middle term will be 14th term
.. As per condition (1)
x13 + x14 + x15 = 177
MEa (0) Q 32 32
substituting from equation 1 we get,
x13 = x1 + 12n
x14 = x1 + 13n
x15 = x1 + 14n
:: x13 + x14 + x15 = (x1+12n) + (x1+13n)
+ (x1+14n) = 177
· (2)
::3x1 + 39n = 177.. (2)
For condition 2, sum of last three
terms = 321
x25 + x26 +x27 = 321
MEg (0) Q 32 32
-(X1 + 24n) + (x1 + 25n) + (x1 + 26n) =
321
: 3x1 + 75n = 321
.. 3x1 = 321 - 75n..
(3)
substituting the value in equation 2,
we get
(321 - 75n) + 39n = 177
. 321 - 177 = 75n - 39n
:: 144 = 36n
.. n = 4. (4)
Substituting in eqn (3),
3x1 + 75x 4 = 321
:: 3x1 + 300 = 321
:: 3x1 = 21
.. x1
= 7 . (5)
1st 3 terms of a.p.
x1 = 7
x2 = x1 + n = 7 + 4 = 11
x3 = x2 + 2 = 11+ 4 = 15
: x1
= 7x2 = 11, x 3 = 15