Find $a_{25} - a_{15}$ for the A.P $- 6, - 10, - 14, - 18, . . . . . . . . . .$

- 40

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- 48

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40

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- 44

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Solution

The correct option is A $- 40$
Given series is: $- 6, - 10, - 14, - 18.....$
$a = - 6, d = - 4$
As we know $n t h$ term, $a_{n} = a + (n - 1) d$
$∴$ $a_{25} = a + 24 d$
& $a_{15} = a + 14 d$
$a_{25} - a_{15} = (a + 24 d) - (a + 14 d) = 10 d$
In the following series $d = - 4$
$∴ a_{25} - a_{15} = 10 \times (- 4) = - 40$

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6^{t h}

17, 14, 11, . . . . . . - 40

Which of the following are isobars?

$_{18}^{40} A r$ $_{6}^{14} C$ $_{20}^{40} C a$ $_{7}^{14} N$

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