Question

3 B Solve.
6
(1) The first term of an A. P. is 1 and the sum of first 15 terms is 225,
find the common difference.​

Answer 1

GIVEN :-

• S_15 = 225.
• First tearm of AP ( a ) = 1.

TO FIND :-

• The common deference ( d ).

SOLUTION :-

As we know that,

$\implies \displaystyle \sf \: S_n = \frac{n}{2} \bigg(2a + (n - 1)d \bigg)$

$\implies \displaystyle \sf \: S_{15} = \frac{15}{2} \bigg(2 + (15 - 1)d \bigg)$

$\implies \displaystyle \sf \: 225 = \frac{15}{2} \bigg(2 + 14d \bigg)$

$\implies \displaystyle \sf \: \frac{225 \times 2}{15} =2 + 14d$

$\implies \displaystyle \sf \: 15 \times 2 = 2 + 14d$

$\implies \displaystyle \sf \: 30 - 2 = 14d$

$\implies \displaystyle \sf \: 28 = 14d$

$\implies \displaystyle \sf \: d = \frac{28}{14}$

$\implies \displaystyle \sf \: d = 2$

Hence the required common deference is 2.

Answer 2

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Understanding \: the \: Question:}}}}}}}$

⋆ This question says that we have to find out the common difference of the Arithmetic progression (AP) whose first term is 1 and the sum of the 15 terms is 225. Let us solve this question!

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Given \: that:}}}}}}}$

⋆ Arithmetic progression is given.

⋆ The first term is 1

⋆ The sum of the 15 terms is 225

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{To \: find:}}}}}}}$

⋆ The common difference

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Solution:}}}}}}}$

⋆ The common difference = 2

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Using \: concept:}}}}}}}$

${\small{\underline{\boxed{\sf{\star \: S_n \: = \dfrac{n}{2} \bigg(2a+(n-1)d\bigg)}}}}}$

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Where...}}}}}}}$

• ${\sf{n \: is \: no. \: of\: terms}}$
• ${\sf{a \: denotes \: AP}}$
• ${\sf{d \: denotes \: common \: difference}}$

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Here...}}}}}}}$

• ${\sf{n \: is \: 15}}$
• ${\sf{S_n \: is \: 225}}$
• ${\sf{a \: is \: 1}}$
• ${\sf{d \: is \: to \: find}}$

${\bigstar \:{\underline{\underline{\pmb{\sf{\purple{Full \: Solution:}}}}}}}$

${\small{\underline{\boxed{\sf{S_n \: = \dfrac{n}{2} \bigg(2a+(n-1)d\bigg)}}}}} \\ \\ :\implies \sf S_n \: = \dfrac{n}{2} \bigg(2a+(n-1)d\bigg) \\ \\ :\implies \sf S_{15} \: = \dfrac{15}{2} \bigg(2(1) + (15-1)d\bigg) \\ \\ :\implies \sf S_{15} \: = \dfrac{15}{2} \bigg(2 + (15-1)d\bigg) \\ \\ :\implies \sf S_{15} \: = \dfrac{15}{2} \bigg(2 + (14)d\bigg) \\ \\ :\implies \sf S_{15} \: = \dfrac{15}{2} \bigg(2+14d\bigg) \\ \\ :\implies \sf 225 \: = \dfrac{15}{2} \bigg(2+14d\bigg) \\ \\ :\implies \sf \dfrac{225 \times 2}{15} \: = 2 + 14d \\ \\ :\implies \sf \dfrac{\cancel{225} \times 2}{\cancel{15}} \: = 2 + 14d \\ \\ :\implies \sf 15 \times 2 \: = 2 + 14d \\ \\ :\implies \sf 30 = 2 + 14d \\ \\ :\implies \sf 30 - 2 \: = 14d \\ \\ :\implies \sf 28 \: = 14d \\ \\ :\implies \sf \dfrac{28}{14} \: = 14d \\ \\ :\implies \sf \cancel{\dfrac{28}{14}} = d \\ \\ :\implies \sf 2 \: = d \\ \\ :\implies \sf d \: = 2 \\ \\ \sf Henceforth, \: common \: difference \: is \: 2$