# Finding a1 and difference in arithmetic sequence

#### eram746

Given: summation of A1 through A10 is 235; and summation of A11 through A20 is 735. Problem asks to find A1 and the difference in this sequence
Thank You

#### skipjack

Forum Staff
Using d for the common difference, you effectively have two equations:

10A1 + 45d = 235
10A1 + 145d = 735

By subtraction, 100d = 500, so d = 5. It follows that A1 = 1.

#### greg1313

Forum Staff
5(2a + 9d) = 235

10(2a + 19d) = 970

2a + 9d = 47

2a + 19d = 97

d = 5

a = 1

#### eram746

5(2a + 9d) = 235

10(2a + 19d) = 970

2a + 9d = 47

2a + 19d = 97

d = 5

a = 1
I am sorry for being slow, but how did you derive to the 2nd equation. I keep getting, 2(2a11+19d) = 147
Edit: are you using the a1+(n-1)d? if so why or Sn= n/2 .....

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#### eram746

Using d for the common difference, you effectively have two equations:

10A1 + 45d = 235
10A1 + 145d = 735

By subtraction, 100d = 500, so d = 5. It follows that A1 = 1.
So, it is valid for me to consider a11 to be a1?
and Sn = 10/2(2a+9d) = 235 -> 10(2a1+9d) = 470 ? no? It would be really helpful you could show the steps for deriving to that answer. Hopefully I am not taking up too much of your time thank you

#### greg1313

Forum Staff
$$S_n=\frac n2(2a+(n-1)d)$$

$$S_{20}=10(2a+19d)=235+735=970$$

O.k?

Sorry about the lack of detail...

1 person

#### aurel5

$$\displaystyle a_1+a_2+a_3+ .. +a_{10} =235 \Rightarrow \dfrac{(a_1+a_{10})10}{2} =235 \Rightarrow (a_1+a_{10})\cdot5=235|_{:5}\Rightarrow a_1+a_{10} = 47 \\\;\\ \Rightarrow a_1+a_1+9d=47 \Rightarrow 2a_1=47-9d\ \ \ (1)$$

$$\displaystyle a_{11}+a_{12}+a_{13}+ .. +a_{20} =735 \Rightarrow \dfrac{(a_{11}+a_{20})10}{2} =735 \Rightarrow (a_{11}+a_{20})\cdot5=735|_{:5}\Rightarrow a_{11}+a_{20} =1 47 \\\;\\ \Rightarrow a_1+10d+a_1+19d=147 \Rightarrow 2a_1=147-29d\ \ \ (2)$$

$$\displaystyle (1), (2) \Rightarrow 47-9d=147-29d \Rightarrow 29d-9d=147-47 \Rightarrow 20d=100\Rightarrow d=5\ \ \ (3) \\\;\\ (1), (3) \Rightarrow 2a_1=47-9\cdot5 \Rightarrow 2a_1=2 \Rightarrow a_1=1. \\\;\\ So,\ a_1=1,\ \ d=5 .$$

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1 person