Question

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

Answer 1

$\bf{\underline{Given}}$

• 3rd term = 12
• last term = 106

$\bf{\underline{To\:find}}$

The 29th term.

$\bf{\underline{Solution}}$

$\sf{\underline{We\:know}}$

$\sf{a_n = a + (n-1)d}$

Therefore,

= $\sf{a_3 = a + (3-1)d}$

$\sf{\implies 12 = a + 2d}$

$\sf{\implies a+2d = 12 \longrightarrow (i)}$

Again,

= $\sf{a_{50} = a + (50-1)d}$

$\sf{\implies 106 = a + 49d}$

$\sf{\implies a + 49d = 106 \longrightarrow (ii)}$

$\sf{\underline{From\:eq.(i)}}$

= $\sf{a+2d = 12}$

$\sf{\implies a = 12-2d}$

$\sf{\underline{Substituting\:the\:value\:of\:a\:in\:eq.(ii)}}$

= $\sf{a+49d = 106}$

$\sf{\implies 12-2d+49d = 106}$

$\sf{\implies 12 + 47d = 106}$

$\sf{\implies 47d = 106-12}$

$\sf{\implies 47d = 94}$

$\sf{\implies d = \dfrac{94}{47}}$

$\sf{\implies d = 2}$

$\sf{\underline{Putting\:the\:value\:of\:d\:in\:eq.(i)}}$

$\sf{a+2d = 12}$

= $\sf{a + 2\times 2 = 12}$

$\sf{\implies a + 4 = 12}$

$\sf{\implies a = 12-4}$

$\sf{\implies a = 8}$

$\sf{\underline{Therefore}}$

$\sf{First\:term\:(a)=8}$

$\sf{Common\:difference\:(d) = 2}$

$\sf{\underline{Now}}$

$\sf{a_{29} = 8 + (29-1)\times2}$

= $\sf{a_{29} = 8 + 28\times2}$

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

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

$\sf{\therefore The\:29th\:term\:is\:64}$

$\bf{\underline{Additional\:Information}}$

$\sf{Sum\:of\:nth\:term\:is\:calculated\:using\:formula:-}$

$\sf{S_n = \dfrac{n}{2}[2a + (n-1)d]}$