Question

The 6th term of an arithmetic sequence is 24. What is the sum of 1st and 11th term of this arithmetic sequence​

Answer 1

$\small \underline \bold{For \: an \: A.P.}$-

$\large \underline \bold{Given}$:-

$\: \: \: \:$$\sf{6th \: term \: = \: A(6) = 24}$

$\large \underline \bold{To \: Find}$:-

$\sf{What's \: the \: sum \: of \: 1st \: and \: 11th \: term \: ?}$

$\large \underline \bold{Using \: identity}$:-

$\: \: \: \: \: \: \: \: \:$$\sf\boxed{\sf\red{an = a + (n - 1)d}}$

$\sf{Here \: ,}$

$\: \: \: \: \: \: \:$$\sf{an = nth \: term}$

$\: \: \: \: \: \: \: \: \:$$\sf{a = first \: term}$

$\: \: \: \: \: \: \: \: \:$$\sf{n = no. \: of \: term}$

$\: \: \: \: \: \: \: \: \:$$\sf{d = Common \: difference}$

$\large \underline \bold{Solution}$:-

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{A(6) = 24}$

$\: \: \:$$\sf{a + (6 - 1)d = 24}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{a + 5d = 24 ---(1)}$

$\sf{and \: ,}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf{A(11) = a + 10d}$

$\small \bold{Now \: ,}$

$\: \: \: \: \: \: \: \: \: \: \: \:$$\sf{A(1) + A(11)}$

$\: \: \: \: \: \: \: \:$$\sf{= a + (a + 10d)}$

$\: \: \: \: \: \: \: \:$$\sf{= 2a + 10d}$

$\: \: \: \: \: \: \: \:$$\sf{= 2(a + 5d)}$

$\: \: \: \: \: \: \: \:$$\sf{= 2(24) \: \: \: \bigg(From \: eq.(1) \bigg)}$

$\: \: \: \: \: \: \: \:$$\sf{= \: 48}$

$\sf\red{Sum \: of \: 1st \: and \: 11th \: term \: of \: this \: A.P. \: is \: 48}$

Answer 2

___________________________

$\fbox\blue{A} \fbox\pink{n} \fbox\orange{s} \fbox\purple{w} \fbox\green{e} \fbox\red{r}$

$\huge \tt Given :-$

• $\: \: \: \: \sf{6th \: term \: = \: A(6) = 24}6thterm=A(6)=24$

$\huge \tt To \: Find :-$

• $\sf{The \: sum \: of \: 1st \: and \: 11th \: term \: }$⁉️

$\: \: \: \: \: \: \: \: \: \sf{\boxed{\sf{\boxed {\pink{an = a + (n - 1)d}}}}}$

$\sf{Here \: ,}$

$\:\implies \sf{a_n = nth \: term}$

$\:\implies \sf{a = first \: term}$

$\: \implies \sf{n = no. \: of \: term}$

$\:\sf{d = Common \: difference}$

$\huge \tt Solution :-$

$\: :\implies \sf{A(6) = 24}A(6)=24$

$\: :\implies \sf{a + (6 - 1)d = 24}a+(6−1)d=24$

$\: :\implies \sf{a + 5d = 24 ---(1)}$

$\sf{and \: ,}$

$\: :\implies \: \sf{A(11) = a + 10d}$

$\small \bold{Now \: ,}$

$\: :\implies \sf{A(1) + A(11)}$

$\: :\implies \sf{a + (a + 10d)}$

$\: :\implies \sf{ 2a + 10d}$

$\: :\implies \sf{2(a + 5d)}$

$\: :\implies \sf{2(24) \: \: \: ...... \: frm\: eq 1 }$

$\: :\implies \sf{\: 48}$

$\tt\green{Sum \: of \: 1st \: and \: 11th \: term \: of \: this \: A.P. \: is \: 48}$✔

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Hope it helps ‼️