Question

the 17th term of an ap is 5 more than twice its 8th term . If the 11 th term is 43 then find a and d​

Answer 1

Given: The 17th Term of an AP is five more than twice the 8th term of the AP. & The 11th term of the AP is 43.

Need to find: The first term ( a ) & the Common difference ( d )?

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☆ By using formula = ⟩⟩ aₙ = a + (n – 1) d ⟨⟨

» 8th term of AP —

$\longrightarrow\bf a_8 = a + 7d\qquad\qquad\quad\sf\Bigg\lgroup eq^{n}\;(1)\Bigg\rgroup$

» 17th term of AP —

$\longrightarrow\bf a_{17} = a + 16d\qquad\qquad\quad\sf\Bigg\lgroup eq^{n}\;(2)\Bigg\rgroup$

$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\; :}}$⠀⠀

• It is Given that, 17th term of the AP is five more than twice the 8th term of the AP.

⠀

$:\implies\sf a_{17} = 2(a_8) +5$

$:\implies\sf a + 16d = 2(a + 7d) + 5\qquad\qquad\quad\sf\Bigg\lgroup Using\;eq^{n}\;(1)\:\&\;eq^{n}\;(2)\Bigg\rgroup$

$:\implies\sf a + 16d = 2a + 14d + 5$

$:\implies\sf 2a - a = 16d - 14d - 5$

$:\implies\bf a = 2d - 5\qquad\qquad\quad\quad\sf\Bigg\lgroup eq^{n}\;(3)\Bigg\rgroup$

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• Second Condition —

• As it is given that, 11th term of the AP is 43.

⠀

$:\implies\sf a_{11} = a + (11 - 1)d$

$:\implies\sf 43 = a + 10d$

$:\implies\sf 43 = 2d - 5 + 10d\qquad\qquad\quad\quad\sf\Bigg\lgroup From\;eq^{n}\;(3)\Bigg\rgroup$

$:\implies\sf 12d = 43 + 5$

$:\implies\sf 12d = 48$

$:\implies\sf d = \cancel\dfrac{48}{12}$

$:\implies{\pmb{\underline{\boxed{\frak{ d = 4}}}}}\;\bigstar$

$\therefore{\underline{\textsf{Hence, the Common difference (d) of the AP is \textbf{4}.}}}$

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✇ Putting the value of d in eqₙ ( 3 ) —

⠀

$:\implies\sf a = 2d - 5$

$:\implies\sf a = 2(4) - 5$

$:\implies\sf a = 8 - 5$

$:\implies{\pmb{\underline{\boxed{\frak{ a = 3}}}}}\;\bigstar$

$\therefore{\underline{\textsf{Hence, the first term (a) of the AP is \textbf{3}.}}}$

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Answer 2

Answer:

Given :-

• The 17th term of an AP is 5 more than twice its 8th term.
• The 11th term is 43.

To Find :-

• What is the value of a and d.

Formula Used :-

$\clubsuit$ General term (nth term) of an AP Formula :

$\mapsto \sf\boxed{\bold{\pink{a_n =\: a + (n - 1)d}}}$

where,

• $\sf a_n$ = nth term of an AP
• a = First term of an AP
• n = Number of terms of an AP
• d = Common difference of an AP

Solution :-

❒ In case of 8th term of an AP :

$\implies \sf a_8 =\: a + (8 - 1)d$

$\implies \sf a_8 =\: a + (7)d$

$\implies \sf\bold{\purple{a_8 =\: a + 7d\: ------\: (Equation\: No\: 1)}}$

❒ In case of 17th term of an AP :

$\implies \sf a_{17} =\: a + (17 - 1)d$

$\implies \sf a_{17} =\: a + (16)d$

$\implies \sf\bold{\purple{a_{17} =\: a + 16d\: ------\: (Equation\: No\: 2)}}$

According to the question,

$\bigstar$ The 17th term of an AP is 5 more than twice its 8th term.

$\implies \sf 17th\: term =\: 2(8th\: term) + 5$

$\implies \sf a + 16d =\: 2(a + 7d) + 5$

$\implies \sf a + 16d =\: 2(a) + 2(7d) + 5$

$\implies \sf a + 16d =\: 2a + 14d + 5$

$\implies \sf 2a - a =\: 16d - 14d - 5$

$\implies \sf a =\: 16d - 14d - 5$

$\implies \sf\bold{\purple{a =\: 2d - 5\: ------\: (Equation\: No\: 3)}}$

Again,

$\bigstar$ The 11th term of an AP is 43.

$\implies \sf a_{11} =\: a + (11 - 1)d$

$\implies \sf a_{11} =\: a + (10)d$

$\implies \sf a_{11} =\: a + 10d$

• 11th term = 43

$\implies \sf 43 =\: a + 10d$

By putting the value of a = 2d - 5 we get,

$\implies \sf 43 =\: (2d - 5) + 10d$

$\implies \sf 43 =\: 2d - 5 + 10d$

$\implies \sf 43 =\: 2d + 10d - 5$

$\implies \sf 43 =\: 12d - 5$

$\implies \sf 43 + 5 =\: 12d$

$\implies \sf 48 =\: 12d$

$\implies \sf \dfrac{48}{12} =\: d$

$\implies \sf \dfrac{4}{1} =\: d$

$\implies \sf 4 =\: d$

$\implies \sf\bold{\red{d =\: 4}}$

Now, we have to find the value of first term of an AP :

Again, by putting the value of d = 4 in the equation no 3 we get,

$\implies \sf a =\: 2d - 5$

$\implies \sf a =\: 2(4) - 5$

$\implies \sf a =\: (2 \times 4) - 5$

$\implies \sf a =\: 8 - 5$

$\implies \sf\bold{\red{a =\: 3}}$

${\footnotesize{\bold{\underline{\therefore\: The\: first\: term\: (a)\: and\: the\: common\: difference\: (d)\: of\: an\: AP\: is\: 3\: and\: 4\: respectively\: .}}}}$