Question

The 2nd,29th and last term of an A.P. are 7 3/4, 1 and -6 1/2 respectively. Find the first term number of terms

Answer 1

Correct question:

$\sf{The \ 2^{nd}, \ 29^{th} \ and \ last \ term \ of \ an \ A.P. } \\ \\ \sf{are \ 7\dfrac{3}{4}, \ 1 \ and \ -6\dfrac{1}{2} \ respectively. } \\ \\ \sf{Find \ the \ first \ term \ and \ number \ of \ terms.}$

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Answer:

$\sf{First \ term \ is \ 8 \ and \ number \ of \ terms \ is \ 55.}$

Given:

$\sf{\leadsto{2^{nd} \ term \ (t_{2})=7\dfrac{3}{4},}} \\ \\ \\ \sf{\leadsto{29^{th} \ term \ (t_{29})=1,}} \\ \\ \\ \sf{\leadsto{n^{th} \ term \ (t_{n})=-6\dfrac{1}{2}}}$

To find:

$\sf{First \ term \ and \ the \ number \ of \ terms.}$

Solution:

$\boxed{\sf{t_{n}=a+(n-1)d}} \\ \\ \\ \textsf{According to the first condition.} \\ \\ \\ \sf{a+d=7\dfrac{3}{4}} \\ \\ \\ \sf{\therefore{a+d=\dfrac{31}{4}...(1)}} \\ \\ \\ \textsf{According to the second condition.} \\ \\ \\ \sf{a+28d=1...(2)} \\ \\ \\ \textsf{Subtract equation (1) from equation (2), we get} \\ \\ \\ \sf{27d=1-\dfrac{31}{4}} \\ \\ \\ \sf{\therefore{27d=\dfrac{-27}{4}}} \\ \\ \\ \sf{\therefore {d=\dfrac{-1}{4}}} \\ \\ \\ \sf{Substitute \ d=\dfrac{-1}{4} \ in \ equation (1), \ we \ get} \\ \\ \\ \sf{a-\dfrac{1}{4}=\dfrac{31}{4}} \\ \\ \\ \sf{\therefore{a=\dfrac{32}{4}}} \\ \\ \\ \boxed{\sf{a=8}} \\ \\ \\ \textsf{According to the third condition. } \\ \\ \\ \sf{-6\dfrac{1}{2}=8+(n-1)\times(\dfrac{-1}{4})} \\ \\ \\ \sf{\dfrac{-11}{2}-8=\dfrac{-1}{4}\times(n-1)} \\ \\ \\ \sf{\therefore{\dfrac{-27}{2}=\dfrac{-1}{4}\times(n-1)}} \\ \\ \\ \sf{\therefore{n-1=54}} \\ \\ \\ \sf{\therefore{n=55}} \\ \\ \\ \purple{\tt{First \ term \ is \ 8 \ and \ number \ of \ terms \ is \ 55.}}$

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

Let assume that the first term and common difference is a and d respectively.

$\rm \Rightarrow \: a + d = 7 \frac{3}{4}$

$\rm \Rightarrow \:a + d = \frac{31}{4}$

$\rm \Rightarrow \:a + (31 - 1)d = \frac{1}{2}$

$\rm \Rightarrow \:a + 30d = \frac{1}{2}$

Solve the above equation

$\rm \Rightarrow \:d = - \frac{1}{4}$

$\rm \Rightarrow \:a = 8$

Assume that the total number of terms are n

$\rm \Rightarrow \:8 + (n - 1) \bigg( - \frac{1}{4} \bigg) = - \frac{13}{2}$

$\rm \Rightarrow \:8 - \frac{1}{4} n + \frac{1}{4} = - \frac{13}{2}$

$\rm \Rightarrow \: \frac{33}{4} - \frac{1}{4} n = - \frac{13}{2 }$

$\rm \Rightarrow \: \frac{1}{4} n = \frac{33}{4} + \frac{13}{2}$

Solve further

$\rm \Rightarrow \: \frac{1}{4} n = \frac{(33 \times 1) + (13 \times 2)}{4}$

$\rm \Rightarrow \: \frac{1}{4} n = \frac{33 + 26}{4}$

$\rm \Rightarrow \: \frac{1}{4} n = \frac{59}{4}$

$\bf \Rightarrow \:n = 59.$