How do I solve this problem 1 by 1 on arithmetic progressions?

Question

In arithmetic progression have 10 terms, the sum of the odd terms is 245. Whereas the sum of the even terms is 305. find the

a)first term
b)common difference

All I know is the formula equation T=a+(n-1)d and S=n/2(2n-1)d

I don't think anyone of you get it right. I tested out your answers but still you all got it wrong. Ok, let's consider this. All of you said a=25 and d=6. Let's see the first ten terms.

25(+6), 31(+6), 37(+6), 43(+6),49(+6), 55(+6), 61(+6), 67(+6), 73(+6) and 79.

Group the odd and even terms and sum it.

Even,
31+43+55+67+79=275
Odd,
25+37+49+61+73=245

You all got the even terms right, but the odd terms wrong.

Answer 1

(a+(n-1)d)+(a+(n-1)d)+(a+(n-1)d)+(a+(n-1)d)+(a+(n-1)d)=S(even)
(a+(1-1)d)+(a+(3-1)d)+(a+(5-1)d)+(a+(7-1)d)+(a+(9-1)d)=305
(a+d)+(a+2d)+(a+4d)+(a+6d)+(a+8d)=305
5a+21d=305...((1))

Odd,
(a+(n-1)d)+(a+(n-1)d)+(a+(n-1)d)+(a+(n-1)d)+(a+(n-1)d)=S(odd)
(a+(2-1)d)+(a+(4-1)d)+(a+(6-1)d)+(a+(8-1)d)+(a+(10-1)d)=245
(a+1d)+(a+3d)+(a+5d)+(a+7d)+(a+9d)=245
5a+25d=245...((2))


Next, use simultaneous equation to solve a and d.

((1)) - ((2))
21d-25d=305-245
-4d=60
d=-15

Substitute d=-15 into ((1)) or ((2))

5a+25d=245
5a+25(-15)=245
5a-375=245
5a=245+375
a=620/5
a=124

Answer 2

you have a slight error on the S equation but no matter.
try this

S = n/2(2a + (n-1)d).

for the ten terms, 5 are odd, 5 are even.

245 = 5/2(2a1 + 4d). = 5a1 + 10d

305 = 5/2(2a2 + 4d). = 5a2 + 10d *by distributive property.

note that they have different starting numbers and a2 - a1 = 2d.

therefore: subtract!
305 - 245 = 5a2 - 5a1 + 10d - 10d
60 = 5(a2 - a1)
divide: 12 = a2 - a1 or 2d = 12, d = 6

for a1 or a...245 = 5/2(2a + 4(12)) => 245 = 5a + 120
125 = 5a
a = 25

hence: first term is 25 and the common diff is 6.

Answer 3

Let's look at odd terms:

S=n/2(2a+(n-1)d)

245=5/2(2a+4d)
5a+10d=245
a+2d=49

For even terms, a becomes a+d

305=5/2(2(a+d)+4d)
5a+15d=305
a+3d=61

Converge the two equations:

49-2d+3d=61
d=12
a=25

BUT d is divided by 2 for original sequence so:

a=25
d=6

Answer 4

sum of odd terms
1,3...9 (5 terms)
this will also form a ap with double common difference
s=5/2(2a+4*(2d) )
245=5/2(2a+8d)
a+4d=49...(1)

sum of even terms
similarly
305=5/2( 2(a+d)+8d)
a+5d=61.....(2)
solve (1) and (2) to get a,d

Answer 5

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