Question

The 17th term of AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find its nth term.

Answer 1

Answer:

$a_{17} = 2a_{8} + 5\\\\a + 16d = 2 ( a + 7d ) + 5\\\\a + 16d = 2a + 14d + 5\\\\a - 2a + 16d - 14d = 5\\\\-a + 2d = 5 \hspace{10mm} => Equation \: 1\\\\a_{11} = a + 10d\\\\a + 10d = 43 \hspace{10mm} => Equation \: 2$

Adding 1 and 2 we get,

2d + 10 d = 5 + 43

=> 12d = 48

=> d = 48 / 12 = 4

=> a + 10d = 43

=> a + 10 ( 4 ) = 43

=> a + 40 = 43

=> a = 43 - 40 = 3

Therefore value of a is 3 and d is 4.

So, the nth term can be given as:

$\implies a_n = a + ( n - 1 ) d\\\\\implies a_n = 3 + ( n - 1 )4\\\\\implies a_n = 3 + 4n - 4 \\\\\implies a_n = 4n - 1$

Hence 4n -1 is the required answer.

Answer 2

Hey there !!

➡ Given :-

→ 17th term of AP is 5 more than twice its 8th term => a $\tiny{17}$ = $2a \tiny{8} \large + 5.$

→ 11th term of an AP is 43.
=> a $\tiny{11}$ = 43.

➡ To find :-

→ nth term ( a $\tiny{n}$ . )

➡ Solution :-

we have,

=> a $\tiny{17}$ = $2a \tiny{8} \large + 5.$

=> a + 16d = 2( a + 7d ) + 5.

=> a + 16d = 2a + 14d + 5.

=> a - 2a + 16d - 14d = 5.

=> -a + 2d = 5.............(1).

And,

=> a $\tiny{11}$ = 43.

=> a + 10d = 43...........(2).

▶ On adding equation (1) and (2), we get

-a + 2d = 5.
a + 10d = 43.
(+)...(+)....(+)
__________

=> 12d = 48.

=> d = $\frac{48}{12} .$

=> d = 4.

▶ Put the value of ‘d’ in equation (2), we get

=> a + 10 × 4 = 43.

=> a + 40 = 43.

=> a = 43 - 40.

=> a = 3.

▶ Now, nth term is given by :-

a $\tiny{n}$ = a + ( n - 1 )d.

=> a $\tiny{n}$ = 3 + ( n - 1 ) × 4.

=> a $\tiny{n}$ = 3 + 4n - 4.

=> $\huge \boxed{ \boxed{ \bf a \tiny{n} \large = 4n - 1. }}$

✔✔ Hence, nth term is 4n - 1. ✅✅

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THANKS

#BeBrainly.