Question

the 11 th term and the 21 st term of an AP are 16 and 29 respectively then the 41 term of that AP

Answer 1

Solution:-

Given

=> T₁₁ = 16

=>T₂₁ = 29

To find

=>T₄₁ =?

Formula

=> Tₙ = a + ( n - 1 )d

Now take

=> T₁₁ = 16

=> 16 = a + ( 11 - 1 )d

=> a + 10d = 16 .......(i)st eq

T₂₁ = 29

=> 29 = a + ( 21 - 1 )d

=> a + 20d = 29 .....(ii)nd eq

Now subtract (i)eq from (ii)eq

=> a + 20d - a - 10d = 29 - 16

=> 10d = 13 => d = 13/10

Now put the value of 10d on first equation

=> a + 10d = 16

=> a + 13 = 16

=> a = 3

Now we have to find T₄₁

=> T₄₁ = a + ( 41 - 1 )d

=> T₄₁ = 3 + 40 × 13/10

=> T₄₁ = 3 + 4 × 13

=> T₄₁ = 3 + 52

=> T₄₁ = 55

Answer 2

Given: The 11th and 21st term of an AP are 16 and 29 respectively.

Need to find: 41th term of that AP.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━

$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\purple{a_n = a + (n - 1)d}}}}$

$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• 11th term of AP is 16.

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a_{11} = a + (11 - 1)d\\ \\ :\implies\sf 16 = a + 10d\\ \\ :\implies\sf a = 16 - 10d\qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

And,

⠀⠀⠀⠀⠀⠀⠀

• 21st term of AP is 29.

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a_{21} = a + (21 - 1)d\\ \\ :\implies\sf 29 = a + 20d\qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

$\dag\;{\underline{\frak{Putting\;eq\:(1)\;in\;eq\:(2),}}}$

$:\implies\sf 29 = (16 - 10d) + 20d\\ \\ :\implies\sf 29 = 16 - 10d + 20d\\ \\ :\implies\sf 29 = 16 + 10d\\ \\ :\implies\sf 10d = 29 - 16\\ \\ :\implies\sf 10d = 13\\ \\ :\implies{\boxed{\sf{\pink{d = \dfrac{13}{10}}}}}\;\bigstar$

☯ Now, Putting value of d in eq (1),

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a = 16 - 10 \bigg( \dfrac{13}{10} \bigg)\\ \\ :\implies\sf a = 16 - 13\\ \\ :\implies{\boxed{\sf{\pink{a = 3}}}}\;\bigstar$

☯ Now, Finding 41th term of AP,

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a_{41} = a + (41 - 1)d$

$\bf Here \begin{cases} & \sf{a = \bf{3}} \\ & \sf{d = \bf{ \frac{13}{10}}} \end{cases}$

$:\implies\sf a_{41} = 3 + 40 \times \bigg( \dfrac{13}{10} \bigg)\\ \\ :\implies\sf a_{41} = 3 + 4 \times 13\\ \\ :\implies\sf a_{41} = 3 + 52\\ \\ :\implies{\underline{\boxed{\sf{\purple{a_{41} = 55}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;41^{th}\;term\;of\;AP\;is\; {\textsf{\textbf{55}}}.}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━

$\boxed{\underline{\underline{\bigstar \: \bf\:Formula\:Related\:to\:AP\:\bigstar}}}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (i)\;The\; n^{th}\;term\;of\;an\;AP\; = \; \red{a_n + (n - 1)d}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (ii)\;Sum\;of\;n\;term\;of\;an\;AP\; = \; \purple{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (iii)\;Sum\;of\;all\;terms\;of\;AP\;having\;last\:term\;as\;'l'\; = \; \pink{ \dfrac{n}{2}(a + l)}$