an ap has 27 terms the sum of the middle term and the two terms adjacent one each side of it is 177 if sum of the last term last three terms is321 find the first three of an ap
Answers
Answer:
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:
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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