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प्रश्न
Find the 19th term of the following A.P.:
7, 13, 19, 25, ...
उत्तर
The sequence given is 7, 13, 19, 25, ....
Here,
First term = a = t1 = 7, t2 = 13, t3 = 19, t4 = 25, ....
Common difference = d = t2 − t1
= 13 – 7
= 6
To find the 19th term, we have to use the formula, i.e.,
tn = a + (n − 1)d
∴ t19 = 7 + (19 − 1) × 6 ...(On substituting value)
∴ t19 = 7 + 18 × 6
∴ t19 = 7 + 108
∴ t19 = 115
Hence, the 19th term of the progression is 115.
संबंधित प्रश्न
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34 + 32 + 30 +...+10
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–12, –5, 2, 9, 16, 23, 30, ..............
For a given A.P. a = 3.5, d = 0, then tn = _______.
If the sum of first n terms of an AP is n2, then find its 10th term.
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Activity :- Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.
Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n − 1) `square`
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`square` = (n – 1) × 5
`square` = (n – 1)
Therefore n = `square`
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Write the next two terms of the A.P.: `sqrt(27), sqrt(48), sqrt(75)`......
