Question

In an arithmetic progression the sum of the first ten terms is 400 and the sum of the next ten terms is 1000. find the common difference and the first term

Answer 1

Step-by-step explanation:

the sum of first 10 terms is S10 while sum of next 10 terms is S20 - S10

Answer 2

Answer:

• Sum of first 10 terms of an AP = 400
• Sum of the next 10 terms of the same AP = 1000
• Common Differce (d) = ?
• First term (a) = ?

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

$\displaystyle\sf :\implies S_{10} = \dfrac{n}{2} \bigg\lgroup 2a+(n-1)d\bigg\rgroup$

$\displaystyle\sf :\implies 400 = \dfrac{10}{2}\bigg\lgroup 2a+(10-1)d\bigg\rgroup$

$\displaystyle\sf :\implies 400 = 5\bigg\lgroup 2a+9d\bigg\rgroup$

$\displaystyle\sf :\implies 400 = 10a+45d\:\:\: -eq(1)$

The sum of the next ten terms

$\displaystyle\sf :\implies S_{20}-S_{10} = \dfrac{n}{2} \bigg\lgroup 2a+(n-1)d\bigg\rgroup$

$\displaystyle\sf :\implies 1000 = \dfrac{20}{2}\bigg\lgroup 2a+(20-1)d\bigg\rgroup - (10a+45d)$

$\displaystyle\sf :\implies 1000 = 10\bigg\lgroup 2a+19d\bigg\rgroup -10a-45d$

$\displaystyle\sf :\implies 1000 = 20a+190d-10a-45d$

$\displaystyle\sf :\implies 1000 = 10a-145d\:\:\:-eq(2)$

• So then we shall do it like elimination method where we'll subtract the two Equations this eliminating one variable and so on rearranging the numbers we'll get one value from which we'll further find the value of the other variable!!

$\displaystyle\sf\dashrightarrow (10a+145d) - (10a+45d) = 1000-400$

$\displaystyle\sf \dashrightarrow 100d = 600$

$\displaystyle\sf \dashrightarrow d = \dfrac{600}{100}$

$\displaystyle\dashrightarrow\underline{\boxed{\sf Common \ Difference = \textsf{ \textbf{6}}}}$

• Now that we have the common differce then the first term a can be found by substituting the value of d in any of the two Equations. So here we shall substitute this in eq(1/ just because the numbers are smaller and makes our calculation easier. This isn't mandatory.

➝ $\displaystyle\sf 400 = 10a+45d$

➝ $\displaystyle\sf 400 = 10a+45\times 6$

➝ $\displaystyle\sf 400 = 10a+270$

➝ $\displaystyle\sf 400-270 = 10a$

➝ $\displaystyle\sf 130 = 10a$

➝ $\displaystyle\sf \dfrac{130}{10} = a$

➝ $\displaystyle\underline{\boxed{\sf First \ term = \textsf{ \textbf{13}}}}$

$\displaystyle\therefore\:\underline{\textsf{C.D will be = \textbf{6} \ \textsf{and the F.T will be \textbf{13}}}}$