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प्रश्न
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
पर्याय
13
9
21
17
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उत्तर
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
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Q.1
Q.11
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
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