Question

An A.P. consist of 50 terms of which third term is 12 and the last term is 106

.Find the 29th term.​

Answer 1

Answer:

Value of 29 th term is 64.

Step-by-step explanation:

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

Then

3rd term= a +(3-1)d= a+2d

50th term= a +(50-1)d=a+ 49d.

Therefore according to the Question,

a+ 2d= 12 ..........(1)

a+ 49d= 106............(2)

Solving (1) and (2) we get:

a= 8 and d = 2.

Therefore, 29 th term

= a+(29-1)d

= 8+28×2

= 8+56

= 64.

Hence the 29 th term is 64

Answer 2

$\boxed{ \red{ \tt{Required \: \: answer :-}}}$

Let the first term be a ; the common difference be d.

${\blue{\tt{A/Q,}}}$

${\blue{\tt{a_3 = 12 \: \: and \: \: a_{50} = 106}}}$

we know that,

${\blue{\tt{A.P = a + (n - 1)d}}}\\{\blue{\tt{↦12 = a + (3 - 1)d}}}\\{\blue{\tt{↦12 = a + 2d \:\:\:\:\: ..(i) }}}\\\\{\blue{\tt{A.P = a+(50 - 1)d}}}\\{\blue{\tt{↦106 = a + (50 - 1)d}}}\\{\blue{\tt{↦106 = a + 49d \:\:\:\:\: ..(ii)}}}</p><p>$

On subtracting eq. (i) from (ii) ,

we get,

${\blue{\tt{↦47d = 94}}}\\{\blue{\tt{↦d = \frac{94}{47}}}}\\\\{\blue{\tt{↦d = \frac{\cancel{94}^2}{\cancel{47}}}}}\\\\{\boxed{\blue{\tt{\therefore d = 2 }}}}$

On putting the value of d in eq.(i),

${\blue{\tt{↦a + 2d = 12}}}\\{\blue{\tt{↦a + 2 × 2 = 12}}}\\{\blue{\tt{↦a = 12 - 4 }}}\\{\boxed{\blue{\tt{\therefore a = 8 }}}}$

so,

${\blue{\tt{\therefore \: 29th\: term = a+(29-1)d}}}\\{\blue{\tt{↦8+28×2 \implies 64 ...(ans)}}}$

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