Question

Solve this arithmetic Progression​

Answer 1

So, here in part a), we are given n terms (no specific no. of terms given), so no. of terms = n , also, the common difference is defined as the difference between the next term to a specific term and the term, i.e ${\bf{d=a_{n}-a_{n-1}}}$, where d is the common difference, so here d = 49 - 45 = 4

Also, here the last term is 49 let say it the nth term, as no number of terms are given, so using the Arithmetic Progression formula, we can write as ;

${:\implies \quad \sf a+(n-1)d=49}$

${:\implies \quad \sf a+(n-1)4=49}$

${:\implies \quad \sf a+4n-4=49}$

${:\implies \quad \sf a=49+4-4n=\boxed{\bf{53-4n}}}$

Now, as we assumed there are n terms, so using the sum formula, we can have ;

${:\implies \quad \sf S_{n}=\dfrac{n}{2}(a+a_{n})}$

${:\implies \quad \sf S_{n}=\dfrac{n}{2}(53-4n+49)}$

${:\implies \quad \sf S_{n}=\dfrac{n}{2}(102-4n)}$

${:\implies \quad \sf S_{n}=n(56-2n)=\boxed{\bf{-2n^{2}+56n}}}$

Now, for part b) we are given a geometric progression with first term = a = 1, common ratio = r = (3/1) = 3, and no. of terms = n = 6, so now putting all values in the sum formula we will have;

${:\implies \quad \sf S_{6}=\dfrac{1\{1-(3)^{6}\}}{1-3}}$

${:\implies \quad \sf S_{6}=\dfrac{1-729}{-2}}$

${:\implies \quad \sf S_{6}=\dfrac{-728}{-2}=\boxed{\bf{364}}}$