Question

an arithmetic sequence has its 5th term equal to 22 and 15th term equal to 62. find its 100th term

Answer 1

d=62-22/15-5
=40/10
=4
T5= a+( n-1)d
22=a+(5-1)(4)
22=a+(4)(4)
22=a+16
a=22-16
a=6

T100= a+(n-1)d
=6+(100-1)(4)
=6+(99)(4)
=6+396
=402

Answer 2

$\huge{\text{\underline{Question}}} -$

An arithmetic sequence has its 5th term equal to 22 and 15th term equal to 62. find its 100th term...???


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$\huge{\text{\underline{Solution}}} -$


$\text{For 5th term}$
an = a + ( n – 1 ) d
a5 = a + ( 5 – 1 ) d
22 = a + 4d
a + 4d = 22 ―――――――――――–―――(1)

$\text{For 15th term}$
an = a + ( n – 1 ) d
a15 = a + ( 15 – 1 ) d
62 = a + 14d
a + 14d = 62 ―――――――――――–―――(2)

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$\text{Subtract eq (1) from (2)}$
$a + 14d = 62 \\ a + \: 4d \: = 22 \\ - \: \: \: \: \: \: - \: \: \: \: \: \: -$
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$\: \: \: \: \: \: \: \: \: \: 10d = 40 \\ = > \: \: \: \: \: d = \frac{40}{10} \\ = > \: \: \: \: \: d = 4$

$\text{Putting the value of d in eq (1)}$

$a + 4d = 22 \\ a + 4 \times 4 = 22 \\ a + 16 = 22 \\ a = 22 - 16 \\ a = 6$

From the Solution, we get

a = 6 and
d = 4


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$\text{For finding 100th term }$

We have to apply nth term formula

So,

an = a + ( n – 1 ) d
a100 = 6 + ( 100 – 1 ) 4
a100 = 6 + 99×4
a100 = 402


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So, the 100th term is 402.


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$\huge{\text{Thanks}}$