Question

14.
The cost of boring a 400-meter deep tube-well keeps increasing with
depth. It is Rs. 10,000 for the first 10 meters and increases by an
additional Rs. 500 for every subsequent 10 meters. Find the cost of
boring the last 10 meters and the total cost of boring.​

Answer 1

Answer:

the answer is

10000+5000= 15000

Answer 2

$\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}$

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$\sf\large\bold{\orange{\underline{\blue{ Given :-}}}}$

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• Depth of tube well is 400 m

• Cost of boring for the first 10 m is Rs 10000

• Cost of boring increases by Rs 500 for every subsequently 10 m

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$\sf\large\bold{\orange{\underline{\blue{ To \: Find :-}}}}$

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• Cost of boring the last 10 m

• Total cost of boring

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$\sf\large\bold{\orange{\underline{\blue{ Solution :-}}}}$

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Given that cost of boring for the first 10 m is Rs 10000 and increases by Rs 500 for every subsequently 10 m

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So ,

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• For first 10 m cost = Rs 10000
• For next 10 m cost = Rs 10500
• For next 10 m cost = Rs 11000

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Clearly the above case is in AP

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10000 , 10500 , 11000 ................ till nth term

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In the question it is given that the tube well need to be bored 400 m deep

The above AP is for consecutive 10 metres hence AP will have :

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➜ $\sf \dfrac { 400 } { 10 }$ Terms

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i.e 40 terms

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$\underline{\bold{\texttt{For nth term :}}}$

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➠ $\sf a_n = a + (n - 1)d$ ⚊⚊⚊⚊ ⓵

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Where ,

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• $\sf a_n$ = nth term

• a = First term

• n = Number of terms

• d = Common difference

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➻ d = 2nd term - 1st term

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$\underline{\bold{\texttt{AP for this case :}}}$

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• $\sf a_n = a_{40}$ Cost of last 10 terms

• a = 10000

• n = 40

• d = 10500 - 10000 = 500

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As each term of this AP signifies the cost of consecutive 10 terms

Thus the last term will give us the cost of boring of last 10 m

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⟮ Putting the above values in ⓵ ⟯

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf a_{40} = 10000 + (40 - 1)500$

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➜ $\sf a_{40} = 10000 + (39)500$

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➜ $\sf a_{40} = 10000 + 19500$

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➨ $\sf a_{40} = 29500$

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• Hence the cost of boring of last 10 m is Rs 29500

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$\underline{\bold{\texttt{Sum of n terms :}}}$

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

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Or,

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➠ $\sf S_n = \dfrac { n } { 2 } (a + l)$ ⚊⚊⚊⚊ ⓶

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Where ,

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• $\sf S_n$ = Sum of n terms

• n = Number of terms

• a = First term

• d = Common difference

• l = Last term

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$\underline{\bold{\texttt{Sum of 40 terms in given AP :}}}$

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• $\sf S_{n}$ = $\sf S_{40}$

• n = 40

• a = 1000

• d = 500

• l = 29500

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➜ $\sf S_{40 }= \dfrac { n } { 2 } (a + l)$

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➜ $\sf S_{40 } = \dfrac { 40} { 2 } (10000 + 29500)$

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➜ $\sf S_{40 } = 20 \times (39500)$

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➨ $\sf S_{40 } = 790000$

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• Hence the cost of boring a 400 m deep tube well is Rs 790000