Question

the eighth term of an AP is 17 and nineteenth term is 39 find the twenty-five term​

Answer 1

51

Step-by-step explanation:

Given

a(8th) = 17

a(19th) = 39

we know that

a(nth) = a + (n-1)d

so,

a(8th) = a + (8-1)d = a + 7d = 17-----1

a(19th) = a + (19 - 1)d = a + 18d = 39-----2

Subtracting 1 and 2 we get

a(19th) - a(8th) = (a + 18d) - (a + 7d) = 39 - 17

= a + 18d - a - 7d = 22

= 11d = 22

= d = 22/11 = 2

Putting d = 2 in eq.1 we get

17 = a + 7(2)

a = 17 - 14 = 3

Now to find the 25th term

a(25th) = 3 + (25 - 1)2

= 3 + 24 × 2 = 3 + 48 = 51

Hope you understood it........All the best

Answer 2

GIVEN :

Eighth term of an AP = 17

nineteenth term = 39

TO FIND :

The twenty-five term of A.P

FORMULA USED :

$Tn = a + (n-1)d$

where,

Tn = nth term

a = first term

n= number of terms

d = common difference

SOLUTION :

$we \: know \: \: Tn = a + (n-1)d$

Eighth term of an AP = 17

That is T8 = 17

Therefore =>

$T8 \: = a + (8-1)d$

$= > 17= a + 7d$

$= > 17 - 7d \: = a$

$= > a = 17 - 7d .......(1)$

nineteenth term = 39

That is T19 = 39

$T19 \: = a + (19-1)d$

$= > 39 \: = a + 18d$

$= > 39 \: - 18d = a$

$= > a = 39 \: - 18d ......(2)$

From equation (1) & (2) :-

=> $17 - 7d = 39 \: - 18d$

=>$17 - 7d + 18d= 39$

=>$- 7d + 18d= 39 - 17$

=>$11d = 22$

=>$d = \frac{22}{11}$

=>$d = 2$

$now \: put \: d = 2 \: in \: a = 17 - 7d$

$a = 17 - (7 \times 2)$

$a = 17 - 14$

$a = 3$

Now to find 25th term , we will use the formula =>

$Tn = a + (n-1)d$

put a = 3 , d = 2 and n = 25

=>$T25 = 3 + (25-1)2$

=>$T25 = 3 + (24 \times 2)$

=>$T25 = 3 + 48$

=>$T25 = 51$

ANSWER :

The twenty-five term of A.P = 51

_____________________________

MORE FORMULAE RELATED TO A.P =>

1. $Tn = a + (n-1)d$

2. Sn = n/2 [ 2a +(n-1)d]

3. Sn = n/2 (a+l)

where,

Tn = nth term

a = first term

n= number of terms

d = common difference

l = last term of A.P