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Help on an arithmetic progression question

The sum of the first ten terms of an AP is 95, and the sum of the first 20 terms of the same AP is 290. Calculate the first term and the common difference.

I've done

S10=5[2a+9d]=95S_{10}=5[2a+9d]=95
S20=10[2a+19d]=290S_{20}=10[2a+19d]=290

Thought about simultaneous equation, tried it but it didn't really come to much..
Reply 1
Avatar for mathz
mathz
8
2a+19d=29
2a+9d=19

dont look the hardest simul equations to solve.
Reply 2
Avatar for Kash:)
Kash:)
OP
15
mathz
2a+19d=29
2a+9d=19

dont look the hardest simul equations to solve.

How did you get the 29 and 19?

From what I can see you divided one by 5 and the other by 10

:confused:
Kash:)
How did you get the 29 and 19?

From what I can see you divided one by 5 and the other by 10

:confused:

That's exactly what he did.
If:
5(2a+9d)=955(2a + 9d) = 95
and
10(2a+19d)=29010(2a + 19d) = 290

Then:
2a+9d=955=192a + 9d = \frac{95}{5} = 19
{lets call this #1}
and
2a+19d=29010=292a + 19d = \frac{290}{10} = 29
{lets call this #2}

Now let's take #1 from #2 and end up with:
10d=1010d = 10

And subsitute d into one of the equations above to find a.
Reply 4
Avatar for Kash:)
Kash:)
OP
15
Wednesday Bass
That's exactly what he did.
If:
5(2a+9d)=955(2a + 9d) = 95
and
10(2a+19d)=29010(2a + 19d) = 290

Then:
2a+9d=955=192a + 9d = \frac{95}{5} = 19
{lets call this #1}
and
2a+19d=29010=292a + 19d = \frac{290}{10} = 29
{lets call this #2}

Now let's take #1 from #2 and end up with:
10d=1010d = 10

And subsitute d into one of the equations above to find a.

Thanks, just saw it :top:

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