Question

An ap consist of 50 terms of which 3rd term is 12 and last term is 106

Answer 1

Answer:the ap is 8,10,12,14...........,106

Step-by-step explanation:

A50=a+49d

106=a+49d........1

A3=a+2d

12=a+2d......2

By elimination. .

a+49d=106

a+2d=12

d=2.....

Puting the value of d in equation 1

I.e. a+2d=12

a+2×2=12

a+4=12

a=12-4

a=8

A2=8+2=10

Hence the following apsara is formed....that is 8,10,12,14.......106

Answer 2

Step-by-step explanation:

${\bf{\blue{\underline{\underline{\: Answer \ :}}}}}$

$\bf{Given}\begin{cases}\sf{ \ 3^{rd} \ term \ is \ 12}\\\sf{ \ Last \ term \ is \ 106}\end{cases}$

Using Arithmetic progression Formula :

$\star\ {\sf{\pink{\underline{\boxed{\sf{a_{n} = a + (n - 1) d}}}}}}$

Ap consist of 50 terms. So, Last term = 50th.

$:\implies\sf a_{l} = a + (50 - 1)d \\\\\\:\implies\sf 106 = a + (50 - 1) d \qquad \qquad \bigg\lgroup\bf a_{l} = 106 \bigg\rgroup\\\\\\:\implies\sf 106 = a + 49d \\\\\\:\implies\sf 106 - 49d = a \\\\\\:\implies\boxed{\sf{\purple{ a = 106 - 49 d }}} \: \sf eq(1)$

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$:\implies\sf a_{3} = a + (3 - 1)d \\\\\\:\implies\sf 12 = a + 2d \qquad \qquad \bigg\lgroup\bf a_{3} = 12 \bigg\rgroup\\\\\\:\implies\sf 12 - 2d = a\\\\\\:\implies\boxed{\sf{\purple{ a = 12 - 2d}}} \: \sf eq(2)$

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$\bigstar$ Now, Finding Common difference (d) from eqn(1) & (2) :

$:\implies\sf a = 106 - 49d \: \& \: a = 12 - 2d \\\\\\:\implies\sf \cancel{ \: a} = 12 - 2d \: \cancel{ = \: a} = 106 - 49d \\\\\\:\implies\sf 12 - 2d = 106 - 49d \\\\\\:\implies\sf -2d + 49d = 106 - 12\\\\\\:\implies\sf 47d = 94\\\\\\:\implies\sf d = \dfrac{\cancel{ 94}}{\cancel{47}}\\\\\\:\implies\boxed{\frak{\purple{ d = 2}}}$

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We get value of d is 2. Let's find out first term (a) of the AP.

$:\implies\sf a = 12 - 2d\\\\\\:\implies\sf a = 12 - 2( 2)\qquad \qquad \bigg\lgroup\bf d = 2 \bigg\rgroup \\\\\\:\implies\sf a = 12 - 4\\\\\\:\implies\boxed{\frak{\pink{a = 8}}}$

Now, we've both values Common difference (d) & First term (a). Finding 31st term of the AP.

$:\implies\sf a_{31} = a + (31 - 1) d \\\\\\:\implies\sf a_{31} = 8 + 30(2)\\\\\\:\implies\sf a_{31} = 8 + 60\\\\\\:\implies\boxed{\frak{\pink{a_{31} = 68}}}$

$\therefore$ Hence, 31st term of the AP is 68.

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⠀⠀⠀⠀⠀⠀⠀⠀$\boxed{\bf{\mid{\overline{\underline{\bigstar\: Formulaes \ : }}}}\mid}$

$\begin{lgathered}\boxed{\begin{minipage}{15 em} \sf \displaystyle \bullet a_n=a + (n-1)d \\\\\\ \bullet S_n= \dfrac{n}{2} \left(a + a_n\right) \end{minipage}}\end{lgathered}$