Question

25th term of an arithmetic sequence is 140 and 27th term is 166 what is common difference and 35th term?

Answer 1

Heya !!



25th term = 140



a + 24d = 140


a = ( 140 - 24d )



And,



27th term = 166


a + 26d = 166


140 - 24d + 26d = 166



2d = 166 - 140


2d = 26



d = 26/2


d = 13.




Therefore,


a = 140 - 24d



=> 140 - 24 × 13


=> 140 - 312


=> -172


Common difference ( d ) = 13.


And,



First term ( a ) = -172



35th term = a + 34d



=> -172 + 34 × 13



=> -172 + 442


=> 270

Answer 2

$\large\underline{Given:-}$

• 25th term of an Ap ⇢ 140
• 26th term of that Ap ⇢ 166

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

• find common different of an Ap ...?
• find it's 15th 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

✰ 25th term of an Ap ⇢ 140

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {({25} \: - \: {1})} \: d \: \: = \: \: {140}$

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {24d} \: \: = \: \: {140} \: \: \: \: \: \: \: ....{(i)}.$

✰ And, 27th term of that Ap ⇢ 166

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {({27} \: - \: {1})} \: d \: \: = \: \: {166}$

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {26d} \: \: = \: \: {166} \: \: \: \: \: \: \: ....{(ii)}.$

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

${a} \: + \: {26d} \: \: = \: \: {166} \\ {a} \: + \: {24d} \: \: = \: \: {140} \\ \underline{- \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: = \: \: \: - \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: {2d} \: \: \: \: \: = \: \: \: {26}$

$\: \: \: \: \: \leadsto \: \: d \: \: = \: \: \frac{26}{2}$

$\: \: \: \: \: \leadsto \: \: {13}$

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

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {24d} \: \: = \: \: {140}$

$\: \: \: \: \: \leadsto \: \: {a} \: + \: {24} \: \times \: {13} \: \: = \: \: {140}$

$\: \: \: \: \: \leadsto \: \: {a} \: - \: {312} \: \: = \: \: {140}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: {140} \: - \: {312}$

$\: \: \: \: \: \leadsto \: \: {a} \: \: = \: \: {-172}$

Therefore,

• first term, a = -172
• common difference, d = 13

✰ Let the 35th term be a + 34d

Then, by putting the value of a and d.

$\: \: \: \: \: \leadsto \:\: {a_{35}} \: \: = \: \: {a} \: + \: {34d}$

$\: \: \: \: \: \leadsto \: \: {a_{35}} \: \: = \: \: {-172} \: + \: {34} \: \times \: {13}$

$\: \: \: \: \: \leadsto \:\: {a_{35}} \: \: = \: \: {-172} \: + \: {442}$

$\: \: \: \: \: \leadsto {a_{35}} \: \: = \: \: {270}$

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

$\: \: \: \: \: \therefore {35th} \: \: term \: \: is \: \: {270}$

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