Question

Maryam gave money to a charity over a 20 year period from year 1 to year 20 inclusive. She gave $150 in year
1, $160 in year 2, $170 in year3, and so on, so that the amount of money she gave each year formed a
sequence. Find the total of money she gave in Year 10?
1
Select one:

a, 1350
b. 1950​

Answer 1

Solution :-

• Money given in first year = $150
• Money given in second year = $160
• Money given in third year = $170

as we can see that, $150,$160,$170 ________ forms an AP sequence .

so,

• First term = a = $150
• common difference = d = 160 - 150 = $10 .
• n = 10

then,

→ S(n) = (n/2)[2a + (n - 1)d]

→ S(10) = (10/2)[2 * 150 + (10 - 1)10]

→ S(10) = 5(300 + 90)

→ S(10) = 5 * 390

→ S(10) = $1950 (Ans.b) .

Hence, the total of money she gave in Year 10 is $1950 .

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Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

Maryam gave money to a charity over a 20 year period from year 1 to year 20 inclusive. She gave $150 in year 1, $160 in year 2, $170 in year3, and so on, so that the amount of money she gave each year formed a

sequence. Find the total of money she gave in Year 10

a, 1350

b. 1950

EVALUATION

Here it is given that Maryam gave money to a charity over a 20 year period from year 1 to year 20 inclusive.

She gave $150 in year 1, $160 in year 2, $170 in year 3, and so on, so that the amount of money she gave each year formed a sequence.

So the given numbers are

150 , 160 , 170 , - - - - - - - ( 20 years)

First term = a = 150

Second Term = 160

Third term = 170

∴ Third term - Second term = Second term - First term

So the above sequence is arithmetic progression

Common Difference = d = 10

We have to find the total of money she gave in Year 10

∴ Time = n = 10 years

So last term of the progression

= 10 th term

= a + ( 10 - 1 ) d

= a + 9d

= 150 + 90

= 240

So the sum of the money after 10 years

$\displaystyle \sf{ = \frac{n}{2} \bigg(First \: term + Last \: term \bigg) }$

$\displaystyle \sf{ = \frac{10}{2}(150 + 240) }$

$= 5 \times 390$

$= 1950$

FINAL ANSWER

Hence the correct option is b. 1950

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