Recent content by Lasher

Oh, that's why))) Thank you))))

Ok..so, there I go \frac{10}{2}[2(12)+(10-1)d]=\frac{15}{2}[2(12)+(15-1)d] - \frac{10}{2}[2(12)+(10-1)d] Therefore 5[24+9d]=7.5[24+14d] - 5[24+9d] Thus 120+45d=180+105d-120+45d So, 45d=180-120+105d+45d-120 45d=60-120+150d -105d=-60 d=-60\div-105 d=\frac{4}{7} There o__O

Seems like math is my worst enemy) What I've done is \frac{10}{2}[2(12)+(10-1)d]=\frac{15}{2}[2(12)+(15-1)d] - \frac{10}{2}[2(12)+(10-1)d]

Thank you so much)) Just to be sure, is the answer S15=240 ?

Ooph.. I get the first part, but now I have trouble finding d. All I get is d=S5-6 Would you be so kind to explain step by step how to find d?

Homework Statement In an arithmetic progression, the sum of the first 10 terms is the same as the sum of the next 5 terms. Given that the first term is 12, find the sum of the first 15 terms. 2. The only one I could think of is S= n/2 (2a+(n-1)d) 3. I've tried solving it, but failed. I...