Question

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then

find its nth term.​

Answer 1

$\\ \\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• 17th term of AP is 5 more than twice its 8th term.

• 11th term of AP is 43.

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• nth term.

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

$\\ \boxed{ \bf \: a_{n} = a + (n - 1)d}$

Here ,

• a(n) = nth term

• a = 1st term

• n = number of terms

• d = common difference

_________________

➣ 11th term , a(11) = a + (11-1)d

➛ a(11) = a + 10d

➛ 43 = a + 10dㅤㅤㅤㅤㅤ--------(i)

_________________

➣ 17th term of AP ,

➛ a(17) = a + (17-1)d

➛ a(17) = a + 16d

➣ 8th term of AP ,

➛ a(8) = a + (8-1)d

➛ a(8) = a + 7d

★ We know , 17th term of AP is 5 more than twice its 8th term.

➛ a(17) = 5 + 2[a(8)]

➛ a + 16d = 5 + 2(a+7d)

➛ a + 16d = 5 + 2a + 14d

➛ a - 2a + 16d - 14d = 5

➛ -a + 2d = 5ㅤㅤㅤㅤㅤ------(ii)

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➣ Adding eqⁿ (i) and (ii) , we get...

➛ a + 10d - a + 2d = 43 + 5

➛ 12d = 48

➛ d = 4

Putting d = 4 in eqⁿ (i),

➛ 43 = a + 10d

➛ 43 = a + 10(4)

➛ 43 = a + 40

➛ a = 43 - 40

➛ a = 3

__________________________

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

➥ a(n) = 3 + (n-1)4

➥ a(n) = 3 + 4n - 4

➥ a(n) = 4n - 1

Hence , nth term is (4n-1).