Question

the mean of 11 items x. if first term is increased by 1, second term by 2 and so on. then new mean becomes ¥ more than x was ¥ is

(1). 4
(2). 8
(3). 6
(4). 12​

Answer 1

6 is the right answer

Answer 2

$\underline{ \underline{ \bold {\red{ \: \: ANSWER\: \: }}}}$➡ $\boxed{ \boxed{ \bold{ \purple{ \: \: (3) \: \: 6 \: \: }}}}$

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$\underline{ \underline{ \bold{ \red{ \: \: SOLUTION\: \:WITH \: \: EXPLAINTION \: \: }}}}$

$\bold{ Given \: \: } \\ \\ \bold{Mean \: \: of \: \: 11 \: \: observation = x} \\ \\ \bold{ = > \frac{ \sum11x}{11} = x} \\ \\ \bold{ = > \sum11x = 11x } \\ \\ \bold{i.e.} \\ \\ \bold{Sum \: \: of \: \: 11 \: \: observation = 11x}$

$\bold{If \: \: first \: \: term \: \: is \: \: increased \: \: by \: \: 1} \\ \bold{second \: \: term \: \: increased \: \: by \: \: 2 \: \: and \: } \\ \bold{so \: \: on \: \: \: it \: \: will \: \: continue \: \:to \: \: 11th \: \: } \\ \bold{term.} \\ \\ \bold{Then, \: \: sum \: \: of \: \: number \: \: 1 \: \: to \: \: 11 \: \: will} \\ \bold{be \: \: added \: \: with \: \: the \: \: sum \: \: of \: \: 11th \: \: } \\ \bold{observations.} \\ \\ \bold{Hence \: ,\: } \\ \\ \bold{The \: \: sum \: \: will \: be = 11x + (1 + 2 + ..... + 11)} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 11x + 66} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 11(x + 6)}$

$\bold{Now,} \\ \\ \bold{The \: \: new \: \: sum \: \: of \: \: the \: \: observation = 11(x + 6)} \\ \\ \bold{No. \: \: of \: \: observation \: = 11} \\ \\ \bold{So,} \\ \\ \bold{Mean = \frac{Sum \: \: of \: \: the \: \: observation}{No. \: \: of \: \: observation} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{11(x + 6)}{11} } \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: = x + 6}$

$\bold{We \: \: get,} \\ \\ \bold{The \: \: old \: \: mean = x} \\ \\ \bold{The \: \: new \: \: mean = x + 6} \\ \\ \bold{Hence,} \\ \\ \bold{The \: \: new \: \: mean \: \: increased = (x + 6) - x} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6}$

$\bold{Thus , }$

$\bold{If \: \: the \: \: mean \: \: of \: \: 11 \: \: items \: \: is \: \: x \: \: and \: } \\ \bold{first \: \: term \: \: increased \: \: by \: \: 1 \: \: second} \\ \bold{term \: \: by \: \: 2 \: \: and \: \: so \: \: on \: \: then \: \: new \: \:} \\ \bold{mean \: \: becomes \: \: \underline{ \: \gamma \: } \: \: more \: \: than \: \: \underline{ \: x \: }.} \\ \\ \bold{Then \: \: the \: \: \underline{ \purple{ \: \: \gamma \: \: }} \: \: is \: \: \underline{ \purple{ \: \: 6 \: \: }} \: \: .}$

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✝$\mathfrak{ \red{ \: \: THANKS...}}$✝