Question

Determine the A P whose third term is 17 and its first term exceeds it's fifth term by 16​

Answer 1

GIVEN :–

• Third term of A.P. = 17

• First term exceeds it's fifth term by 16.

TO FIND :–

• Determine the A.P. = ?

SOLUTION :–

• We know that nth term of A.P. –

$\\ \implies \bf T_n=a+(n-1)d$

• According to the first condition –

$\\ \implies \bf T_3=17$

$\\ \implies \bf a+(3-1)d=17$

$\\ \implies \bf a+2d=17$

$\\ \implies \bf a=17 - 2d \: \: \: \: \: \: \: \: - - - eq.(1)$

• According to the second condition –

$\\ \implies \bf T_1 = T_5 + 16$

$\\ \implies \bf a+(1-1)d=a+(5-1)d+ 16$

$\\ \implies \bf a=a+4d+ 16$

$\\ \implies \bf 4d+ 16 = 0$

$\\ \implies \bf 4d = - 16$

$\\ \implies \large{\boxed{ \bf d =-4}}$

• Put in eq.(1) –

$\\ \implies \bf a=17 - 2( - 4)$

$\\ \implies \bf a=17 + 8$

$\\ \implies \large{ \boxed{ \bf a=25}}$

• Hence , The series is 25,21,17,13,. . . . . . . .

Answer 2

Answer:

Required Answer :-

As we know that

ⁿth term = a + (n - 1) d

$\sf T_3$ = 17

17 = a+ (3 - 1)d

17 = a + 2d

a = 17 - 2d (Eqⁿ 1)

T_1 = T_5 + 16

a + (1 - 1)d = a + (5 - 1)d + 16

a = a + (5 - 1)d + 16

a = a + 4d + 16

4d = -16

d = -4

Let's put the value in Eqⁿ 1

a = 17 - 2(-4)

a = 17 + 8

a = 25