Find the sum of the first 100 terms of the arithmetic sequence with the first term 2 and the common difference 5.
First term==2
Common difference ==5
The sequence is:
2 , 7 , 12 , 17 , 22 , 27 , 32 , 37 , 42 , 47 , 52 , 57 , 62 , 67 , 72 , 77 , 82 , 87 , 92 , 97 , 102 , 107 , 112 , 117 , 122 , 127 , 132 , 137 , 142 , 147 , 152 , 157 , 162 , 167 , 172 , 177 , 182 , 187 , 192 , 197 , 202 , 207 , 212 , 217 , 222 , 227 , 232 , 237 , 242 , 247 , 252 , 257 , 262 , 267 , 272 , 277 , 282 , 287 , 292 , 297 , 302 , 307 , 312 , 317 , 322 , 327 , 332 , 337 , 342 , 347 , 352 , 357 , 362 , 367 , 372 , 377 , 382 , 387 , 392 , 397 , 402 , 407 , 412 , 417 , 422 , 427 , 432 , 437 , 442 , 447 , 452 , 457 , 462 , 467 , 472 , 477 , 482 , 487 , 492 , 497 , >>Number of terms = 100>>Total Sum == 24950
+26388 Find the sum of the first 100 terms
of the arithmetic sequence
with the first term 2
and the common difference 5.
\(\begin{array}{|rcll|} \hline \text{sum} &=& \dfrac{2+\Big( 2+(100-1)*5 \Big)}{2}*100 \\\\ \text{sum} &=& (2+2+99*5)*50 \\\\ \text{sum} &=& (4+495)*50 \\\\ \text{sum} &=& 499*50 \\\\ \mathbf{\text{sum}} &=& \mathbf{24950} \\ \hline \end{array}\)
