Question

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term
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Answer 1

$\sf\huge\blue{\underline{\underline{ Question : }}}$

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• An AP series consists of 50 terms. (n)
• 3rd term (a3) = 12
• Last term (an) = 106.

★ To find,

• 29th term.

★ Formula used :

• $\tt\green{:\implies a_{n} = a + (n - 1)d }$

★ Let,

$\sf\:\implies a_{3} : a + 2d = 12 ..... (1)$

Last term of AP is 106. (an)

• Substitute the values in the formula.

$\sf\:\implies 106 = a + (50 - 1)d$

$\sf\:\implies a + 49d = 106 ..... (2)$

Subtract equations (1) & (2). We get,

$\sf\:\implies -2d = -94$

$\sf\:\implies d =\frac{-94}{-2}$

$\sf\:\implies d = 2$

• Substitute the value of d in (1)

$\sf\:\implies a + 2(2) = 12$

$\sf\:\implies a + 4 = 12$

$\sf\:\implies a = 12 - 4$

$\sf\:\implies a = 8$

★ Now,

We need to find the 29th term.

• $\tt\red{:\implies a_{29} = a + 28d }$
• Put the values of a & d

$\sf\:\implies a_{29} = 8 + 28(2)$

$\sf\:\implies a_{29} = 8 + 56$

$\sf\:\implies a_{29} = 64$

$\underline{\boxed{\bf{\purple{ \therefore 29^{th} \:term\:of\:the\:AP = 64.}}}}\:\orange{\bigstar}$

★ More information :

How to find common difference (d)?

Let us take an AP series.

AP : 1,2,3....

Let,

$\tt\:\implies a_{1} = 1$

$\tt\:\implies a_{2} = 2$

$\tt\:\implies a_{3} = 3$

Common difference (d) : a2 - a1 = a3 - a2

$\sf\:\implies 2 - 1 = 3 - 2$

$\sf\:\implies 1 = 1$

↪ From the above we can see that the difference between the successive terms is same (constant) which is 1.

↪ so we can say that the given sequence is in A.P.

↪ If the 1st term and the common difference 'd' is given then we can make an arithmetic sequence.

___________________________

$\boxed{\begin{minipage}{5 cm} AP Formulae \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

___________________________

Answer 2

64

Step-by-step explanation:

Let the first term of the AP be 'a' and common difference be 'd'.

3rd term = 12    ⇒ a + 2d = 12     ...(1)

50th term = 106  ⇒ a + 49d = 106   ...(2)

Subtract (1) from (2),  we get 47d = 94

         ⇒ d = 94/47

         ⇒ d = 2

Hence, in (1),   a + 2(2) = 12    ⇒ a = 8

∴ 29th term = a + 28d

                   = 8 + 28(2)

                   = 64