Question

The 7th term of an ap is -4 and its 13th term is -16.Find the first term, common difference and nth term

Answer 1

Answer:

♣️Refer To Attachment♣️

I hOpe It HeLpS Ya!

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Answer 2

$\large\underline{Given:-}$

• 7th term of an Ap ⇢ -4
• 13th term of that Ap ⇢ -16

$\large\underline{To find:-}$

• find the first term, common different and nth term ...?

$\large\underline{Solutions:-}$

$\: \: \: \: \: \therefore \: \: \: {a_n} \: \: = \: \: {a} \: + \: {(n \: - \: {1})} \: d$

• a ⇢ first term
• d ⇢ common difference
• n ⇢ number of term
• an ⇢ last term

»★ We have

✰ 7th term of an Ap ⇢ -4

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {({7} \: - \: {1})} \: d \: \: = \: \: {-4}$

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {6d} \: \: = \: \: {-4} \: \: \: \: \: \: \: ....{(i)}.$

✰ And, 13th term of that Ap ⇢ -16

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {({13} \: - \: {1})} \: d \: \: = \: \: {-16}$

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {12d} \: \: = \: \: {-16} \: \: \: \: \: \: \: ....{(ii)}.$

Now, Subtracting Eq. {(i)} and Eq. {(i)}

${a} \: + \: {6d} \: \: = \: \: {-4} \\ {a} \: + \: {12d} \: \: = \: \: {-16} \\ \underline{- \: \: \: \: \: \: \: \: - \: \: \: \: \: = \: \: \: \: \: \: \: + \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: {-6d} \: \: \: \: \: = \: \: \: {12}$

$\: \: \: \: \: \leadsto \: \: d \: \: = \: \: \frac{12}{-2}$

$\: \: \: \: \: \leadsto \: \: {-2}$

»★ Now, putting the value of d in Eq. (i)

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {6d} \: \: = \: \: {-4}$

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {6} \: \times \: {-2} \: \: = \: \: {-4}$

$\: \: \: \: \: \leadsto \: \: {a} \: - \: {-12} \: \: = \: \: {-4}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: {-4} \: + \: {12}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: {8}$

Therefore,

• first term, a = 8
• common difference, d = -2

✰ Let the nth term be a + (n - 1) d

Then, by putting the value of a and d.

$\: \: \: \: \: \leadsto \: \: {a_{n}} \: \: = \: \: {a} \: + \: {( n \: - \: 1)} \: d$

$\: \: \: \: \: \leadsto \: \: {8} \: + \: {( {n} \: - \: 1)} \: {-2}$

$\: \: \: \: \: \leadsto \: \: {8} \: + \: {-2n} \: - \: {2}$

$\: \: \: \: \: \leadsto \: \: {-2n} \: + \: {10}$

$\: \: \: \: \: \leadsto \: \: {2n} \: \: = \: \: {10}$

$\: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: \frac{10}{2}$

$\: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: {5}$

$\: \: \: \: \: \: \: \: Hence,$

$\: \: \: \: \: \therefore {nth} \: \: term \: \: is \: \: {5}$

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