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प्रश्न
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference.
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उत्तर
Let a and d be the first term and the common difference of the AP, respectively.
∴ nth term of the AP, an = a + (n − 1)d
Given:
\[\Rightarrow a + \left( 4 - 1 \right)d = 11\]
\[ \Rightarrow a + 3d = 11 . . . . . \left( 1 \right)\]
Also, \[a_5 + a_7 = 34\]
\[\Rightarrow \left( a + 4d \right) + \left( a + 6d \right) = 34 \]
\[ \Rightarrow 2a + 10d = 34\]
\[ \Rightarrow a + 5d = 17 . . . . . \left( 2 \right)\]
Solving (1) and (2), we get
a = 2 and d = 3
Hence, the common difference of the AP is 3.
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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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