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प्रश्न
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
उत्तर
It is given that,
t4 = –15
t9 = –30
Now,
tn = a + (n - 1)d
t4 = a + (4 - 1)d
⇒ -15 = a + 3d
⇒ a = -15 - 3d ...(1)
⇒ t9 = a + (9 - 1)d
⇒ -30 = a + 8d
⇒ a + 8d = -30
⇒ -15 - 3d + 8d = -30 ...(From 1)
⇒ -15 + 5d = -30
⇒ 5d = -30 + 15
⇒ 5d = -15
⇒ d = -3
⇒ a = -15 - 3(-3) ...(From 1)
⇒ a = -15 + 9
⇒ a = -6
t1 = a = -6
t2 = t1 + d = -6 - 3 = -9
t3 = t2 + d = -9 - 3 = -12
t4 = t3 + d = -12 - 3 = -15
Hence, the given A.P. is –6, –9, –12, -15, ....
Now,
\[S_n = \frac{n}{2}\left( 2a + \left( n - 1 \right)d \right)\]
\[ S_{10} = \frac{10}{2}\left( 2a + \left( 10 - 1 \right)d \right)\]
= 5 (2 (-6) + 9 (-3))
= 5 (-12 - 27)
= 5 (-39)
= -195
Hence, the sum of the first 10 numbers is –195.
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Q.18
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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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