Question

the seventeenth term of an ap is 5 morethan twice its eighteenth term . If the 11 term of the ap is 43 find its nth term

Answer 1

Answer:

nth term = 109 - 6n

or

103 - 6(n-1)

Step-by-step explanation:

nth term of an ap = a + (n-1)d

17th Term = a + (17-1)d   = a + 16d

18th term = a + (18-1)d = a + 17d

17th term = 2 * (18th term) + 5

a + 16d = 2 * ( a+ 17d) +5

a +16d = 2a + 34d + 5

a + 18d = - 5    Eq 1

11th term = a + (11-1)d = a + 10d

a + 10d = 43     Eq 2

Eq 1 - Eq 2

8d = -48

d = -6

a + 10(-6) = 43

a = 103

nth rem = 103+ (n-1)(-6)

= 103 - 6(n-1)

= 103 - 6n + 6

= 109 - 6n

Answer 2

a17 = a+16 d

a18 = a+17 d

$\textbf{\underline{Given that :-}}$

a17 = 5 + 2 × a18

a+16d = 5+ 2(a+17 d)

a+16 d = 5+2a+34d

16d-34d -5 =2a-a

$\boxed{ \bf{ - 18 \: d - 5 = a}}$

a11 =43

a+10d = 43

-18d -5 +10d =43

-8d -5 =43

-8d =43+5

-8d =48

$\bf{d = \frac{48}{ - 8} }$

$\boxed{ \bf{d = - 6}}$

a=-18 d -5

a= -18×-6 -5

a= 108 -5

$\boxed{ \bf{a = 103}}$

$\underline{ \bf{a {\tiny{n}} \: = \: a + (n - 1)d }}$

an = 103 + (n-1) × -6

an =103 -6n +6

$\boxed{ \bf{a{ \tiny{n}} = 109 - 6n}}$