hello

here's the question I need a hand with, it's about arithmetic progressions:

Find the first three terms of an arithmetic sequence where
T5 (term 5)= 3xT2 and the sum of the sequence to six terms is 36

any help is much appreciated so thanks in advance

Tn = a + (n-1)d
where a denotes first term and d denotes the common difference ( difference between any term and the previous term)

so T2 = a + d
T5 = a + 4d

but T5 = 2*T2

so a+4d = 2(a+d)

which simplifies to a = 2d

Sn the sum of n terms is given by

Sn = n/2[2a+ (n-1)d]

so S6 = 6/2[2a+5d] = 36

so 6a + 15d = 36

but a= 2d from above

so 12d + 15 d = 27d = 36

d = 4/3

a = 2d = 8/3

thanks so much for the help

sorry!

misread question

T5 = 3T2

so a+4d = 3a+3d

Can you take it from here?

Arithmetic sequence:
Tn = T1 +(n -1)d
Sn = (n/2)[T1 +Tn]

T5 = 3(T2) .....given.
T1 +(5 -1)d = 3[T1 +(2 -1)d]
T1 +4d = 3T1 +3d
4d -3d = 3T1 -T1
d = 2T1 ---------------(1)

S6 = 36 ...........given.
S6 = (6/2)[T1 +Tn] = 36
3[T1 +(T1 +(6-1)d)] = 36
3[T1 +T1 +5d] = 36
6T1 +15d = 36
2T1 +5d = 12
d = (12 -2T1)/5 -------(2)

d = d,
2T1 = (12 -2T1)/5
10T1 = 12 -2T1
12T1 = 12
T1 = 1 ----------**

So, d = 2T1 = 2(1) = 2

Therefore,
T1 = 1 -------------------answer.
T2 = 1 +2 = 3 ------------answer.
T3 = 1 +2(2) = 5 ----------answer.

Check:
T6 = 1 +5(2) = 11
S6 = (6/2)[1 +11] = 3(12) = 36 .....OK.

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