Question

solve it by step by step explanation

Answer 1

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

$\[\]\\ \begin{equation}a_p=a+(p-1)d=q\end{equation}\\ \\ \begin{equation}a_q=a+(q-1)d=p\end{equation}$

By subtracting (2) from (1),

$\begin{aligned}&(1)-(2)\\ \\ \Longrightarrow\ \ &(a+(p-1)d)-(a+(q-1)d)=q-p\\ \\ \Longrightarrow\ \ &a+(p-1)d-a-(q-1)d=q-p\\ \\ \Longrightarrow\ \ &(p-1-q+1)d=q-p\\ \\ \Longrightarrow\ \ &(p-q)d=q-p\\ \\ \Longrightarrow\ \ &d=-1\end{aligned}$

Adding (1) and (2),

$\begin{aligned}&(1)+(2)\\ \\ \Longrightarrow\ \ &a+(p-1)d+a+(q-1)d=p+q\\ \\ \Longrightarrow\ \ &2a-1(p-1+q-1)=p+q\ \ \ \ \ [\because\ d=-1]\\ \\ \Longrightarrow\ \ &2a-(p+q-2)=p+q\\ \\ \Longrightarrow\ \ &2a-p-q+2=p+q\\ \\ \Longrightarrow\ \ &2a+2=2p+2q\\ \\ \Longrightarrow\ \ &a+1=p+q\\ \\ \Longrightarrow\ \ &a=p+q-1\end{aligned}$

Now,

$a_n=a+(n-1)d\\ \\ a_n=p+q-1-(n-1)\\ \\ a_n=p+q-1-n+1\\ \\ \mathbf{a_n=p+q-n}$

Answer 2

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

□Let first term be=a

□Common difference(c.m)=d

$\huge{\boxed{\boxed{\underline{\sf{TO\:PROVE-}}}}}$

☆an=p+q-n

$\huge{\boxed{\boxed{\underline{\sf{PROOF:}}}}}$

$ap = a + (p - 1) \times d = q - - - - - (1) \\ aq = a + (q - 1) \times d = p \\ first \: subtract \: (2) \: from \: (1) we \: get \\ = ) a + (p - 1)d - (a + (q - 1)d) = q - p \\ = )a + (p - 1)d - a - (q - 1)d = q - p \\ = )(p - 1)d - (q - 1)d = q - p \\ = )(p - 1 - q + 1)d = q - p \\ = )(p - q)d = q - p \\ = )d = - 1 - - - - - (3)\\ > > so \: again \: we \: ad ding\: (1) \: and \: (2) \\\\ = )a + (p - 1)d + a + (q - 1)d = p + q \\ > > putting \: value \: of \: d = - 1 \\2a - (p + q 1 - 1) = p + q \\ = )2a - (p + q - 2) = p + q \\ = )2a - p - q + 2 = p + q \\ = )2a + 2 = 2p + 2q \\ = )2(a + 1) =2( p + q) \\ = )a = p + q - 1 - - - - - (4) \\ so \: again \\ an = a + (n - 1)d \\ > > putting \: value \: of \: a \: and \: from \: (3) \: and \: (4) \\ \\an = p + q - 1 - (n - 1) \times 1 \\ an = p + q - 1 - n + 1 \\ an = p + q - n$

$\huge{\boxed{\boxed{\underline{\sf{PROVED-}}}}}$

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