Question

For an A.P., the 6th term is 19 and the 17th term is 41. Find the 40th term of the A. P.

Answer 1

GIVEN

6th Term of the AP = 19

17th Term of the AP = 41

TO FIND :- 40th Term of the AP

SOLUTION

a + 5d = 19 ... eq.01

a + 16d = 41 ...eq.02

Subtracting eq.01 from eq.02;

a + 16d - a - 5d = 41 - 19

=>11d = 22

=> d = 2

Substituting value in eq.01,

a + 5 × 2 = 19

=> a = 19 - 10

=> a = 9

40th Term = a + 39d

= 9 + 39 × 2

= 9 + 78

= 87 (ANS)

Answer 2

$\large{ \bold{ \fbox{ \fbox{Given :- \: }}}}$

$\to \bold{ {17}^{th} \: Term \: Of \: The \: AP = 41 \: } \\ \to \bold{ {6}^{th} \: Term \: Of \: The \: AP = 19}$

$\large{ \bold{ \fbox{ \fbox{To \: \: Find :- \: }}}}$

$\to\bold{ {40}^{th} \: Term \: Of \: The \: AP = ? \: }$

$\large{ \bold{ \fbox{ \fbox{Solution :- \: }}}}$

$\to \bold{a + 16d = 41 - - - - - \: ( i ) } \\ \\ \to\bold{a + 5d = 19 - - - - - \: (ii) }</p><p>$

$\large{ \bold{ \underline{Subtract \: Equation \: ( i ) \: And \: ( ii )</p><p> \: \: \: }}}$

$\to \bold{( a + 16d ) - ( a + 5d ) = 41 - 19 \: } \\ \\ \to\bold{a + 16d - a - 5d = 22 \: } \\ \\ \to \bold{11d = 22 \: } \\ \\ \to \bold{d = \frac{22}{11} } \\ \\ \to\bold{d = 2}$

$\large{ \bold{ \underline{Put \: The \: Value \: of \: d = 2 \: In \: Equation \: ( ii ) \: \: \: }}}$

$\to \bold{a + 5 × 2 = 19 \: } \\ \to\bold{a + 10 = 19 \: } \\ \to \bold{a = 19 - 10 \: } \\ \to \bold{a = 9 \: }$

We Have To Find 40^th Term Of The AP So ,

$\to \bold{ {40}^{th} \: Term \: Of \: The \: AP = a + 39d \: }</p><p> \\ \\ \to \bold{ {40}^{th} \: Term \: Of \: The \: AP = 9 + 39 × 2 \: } \\ </p><p> \\ \to \bold{ {40}^{th} \: Term \: Of \: The \: AP = 87 \: }$