Question

The 10th term of an ap is 20 and 20th term is 10 find 30th term.

Answer 1

Answer:

answer is 30th term is 5find

good morning have a great day

Answer 2

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$\large\underline{Given:-}$

• 10 times 10th term is equal to 20 time 20th term.

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

• find it's 30th term...?

$\large\underline{Solutions:-}$

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

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

»★ We have

✰ 10th term

$\: \: \: \: \: \leadsto {a_{10}} \: \: = \: \: {a} \: + \: {({10} \: - \: {1})} \: d$

$\: \: \: \: \: \leadsto {10} \: \: = \: \: {a} \: + \: {9d} \: \: \: \: \: \: \: ....{(i)}.$

✰ And, 20th term

$\: \: \: \: \: \leadsto {a_{10}} \: \: = \: \: {a} \: + \: {({20} \: - \: {1})} \: d$

$\: \: \: \: \: \leadsto {20} \: \: = \: \: {a} \: + \: {19d} \: \: \: \: \: \: \: ....{(ii)}.$

»★ Now, solving Eq. (i)and Eq. (ii)

$\: \: \: \: \: \leadsto {10} \: {({a} \: + \: {9d})} \: \: = \: \: {20} \: {({a} \: + \: {19d})}$

$\: \: \: \: \: \leadsto {10a} \: + \: {90d} \: \: = \: \: {20a} \: + \: {380d}$

$\: \: \: \: \: \leadsto {20a} \: - \: {10a} \: \: = \: \: {90d} \: - \: {380d}$

$\: \: \: \: \: \leadsto \: \: {10a} \: \: = \: \: {-290d}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: \frac{-290d}{10}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: {-29d}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: \frac{-290d}{10}$

✰ Let the 30th term be a + 29d

Then, by putting the value of a.

$\: \: \: \: \: \leadsto {a_{30}} \: \: = \: \: {a} \: + \: {29d}$

$\: \: \: \: \: \leadsto {a_{30}} \: \: = \: \: {-29d} \: + \: {29d}$

$\: \: \: \: \: \leadsto {a_{30}} \: \: = \: \: {0}$

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

$\: \: \: \: \: \therefore {30th} \: \: term \: \: is \: \: {0}$

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